Sequences as functions whose domain is a set of consecutive integers, writing terms from a rule and a rule from terms, graphing a sequence as isolated points, series and summation notation, and the three special formulas for sums.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 12 — Sequences and Series
Define and Use Sequences and Series
Objectives
Five outcomes. A list with a rule, and the total when you add it up.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 794-801 — the lesson these objectives are drawn from
Warm-up
You have worked with functions since Chapter 2, taking an input to an output.
Discussion prompt
The list 2, 4, 6, 8 pairs the first position with 2, the second with 4, and so on. Is that a function? What is its domain?
Hint: What goes in, and what comes out?
Answer:
It is a function. The input is the POSITION — first, second, third — and the output is the term at that position.
\[ f(1) = 2, \; f(2) = 4, \; f(3) = 6, \; f(4) = 8 \;\Longrightarrow\; f(n) = 2n \]
A sequence is exactly that: a function whose domain is a set of consecutive integers. Everything you know about functions applies, with the input restricted to whole numbers.
Concept
A sequence is a function whose domain is a set of consecutive integers, usually starting at 1. Its outputs are the terms. Adding the terms together gives a series, written compactly in summation notation.
sequence — A function whose domain is a set of consecutive integers. Its outputs are called the terms. A finite sequence has a limited number of terms; an infinite one continues without stopping.
\[ a_n = 2n; \qquad \sum_{i=1}^{4} 2i = 20 \]
The notation a sub n and f of n mean the same thing. The subscript form is traditional for sequences and emphasises that n counts positions.
Figure (svg): Two columns comparing a sequence with a series
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 794-796
Section
Section 1
Concept
A sequence's rule is a formula for the term at position n. Substituting n equal to 1, 2, 3 and so on generates the terms in order.
\[ a_n = 2n+5 \;\Longrightarrow\; 7, 9, 11, 13, \dots \]
Unless a domain is stated the positions start at 1, so a sub 1 is the first term rather than a sub 0.
Figure (svg): A sequence shown as a function pairing each position with a term
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 794-794 — Write terms of sequences
Picture it
Example 1a: the rule 2n plus 5.
Figure (svg): A sequence shown as a function pairing each position with a term
Each position on the left maps to one term on the right. The rule is the arrow, and the list of outputs is the sequence.
Worked example
Example 1, both parts.
\[ \text{Write the first six terms of } a_n = 2n+5 \text{ and } f(n) = (-3)^{n-1}. \]
First rule at n equal to 1 and 2
Why: Two plus 5, then 4 plus 5.
\[ 7\text{ and } 9 \]
Continue to n equal to 6
Why: Each step adds 2.
\[ 7, 9, 11, 13, 15, 17 \]
Second rule at n equal to 1
Why: Negative 3 to the power zero.
\[ 1 \]
Continue
Why: Each step multiplies by negative 3.
\[ 1, -3, 9, -27, 81, -243 \]
Figure (svg): A sequence shown as a function pairing each position with a term
\[ 7,9,11,13,15,17; \quad 1,-3,9,-27,81,-243 \]
Verify: check the exponent's offset
Why: The second rule uses n minus 1, so the first term has exponent 0 and equals 1 rather than negative 3. Off-by-one errors in an exponent are the commonest slip here, and testing n equal to 1 first catches them immediately.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 794-794
Fill the middle
Example 1b.
Fill in the blanks
f(1) = (-3)^1 = (-3)^0 = ___
Why: Any nonzero base to the power zero is 1, so the first term is 1 rather than negative 3. The n minus 1 in the exponent is what shifts the powers.
Worked example
Guided Practice 1 to 3.
\[ \text{Write six terms of } a_n = n+4, \; f(n) = (-2)^{n-1}, \; a_n = \tfrac{n}{n+1}. \]
First: add 4 to each position
Why: One plus 4, 2 plus 4, and so on.
\[ 5, 6, 7, 8, 9, 10 \]
Second: powers of negative 2
Why: Starting at the zeroth power.
\[ 1, -2, 4, -8, 16, -32 \]
Third: position over position plus one
Why: One over 2, 2 over 3, and so on.
\[ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7} \]
Note the third's behaviour
Why: The fractions creep toward 1 without reaching it.
Figure (svg): The solution to Worked example three more rules shown as a ladder of expressions, one row per algebraic move
\[ 5,6,\dots; \; 1,-2,4,\dots; \; \tfrac{1}{2},\tfrac{2}{3},\tfrac{3}{4},\dots \]
Verify: check the third's limit
Why: The terms are n over n plus 1, which is always less than 1 and always increasing — at n equal to 100 it is 100 over 101, about 0.99. The sequence approaches 1 but never reaches it, which is the same asymptote behaviour as the rational functions of Lesson 8.2.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 794-794
Trap
\[ a_n = 2n+5 \]
Take the first term as a sub 0
Why: The habit of starting counts at zero is carried over.
\[ a_0 = 5 \quad \text{(not the first term)} \]
Unless a domain is stated, a sequence's positions start at 1, so the first term is a sub 1, which is 7.
\[ a_1 = 2(1)+5 = 7 \]
Start the positions at 1
Why: Position 1 is the first term, position 2 the second, and so on.
\[ 7, 9, 11, 13, 15, 17 \]
Some sequences are deliberately defined from n equal to 0, but the definition says so. When nothing is said, count from 1.
Matching
Substitute 1, 2 and 3.
Match the pairs
Why: The first and third both add a constant amount each step, the second multiplies, and the fourth does neither. Those three behaviours are the arithmetic, geometric and other cases the next two lessons will separate.
Sorting
Does the list stop?
Sort into buckets
Sort each sequence.
The three dots do the work in written lists, and the situation does it in real problems — a pyramid has as many layers as it has and no more.
Prediction
Commit before reasoning.
Predict first
What makes a sequence a function rather than just a list?
Correct: Each position has exactly one term, so the positions are inputs and the terms are outputs.
\[ \text{domain } 1,2,3,\dots; \quad \text{range } a_1, a_2, a_3, \dots \]
Why: The defining property of a function is one output per input, and a list gives exactly one third term, one fourth term and so on. Calling it a function means the whole toolkit applies: a rule, a domain, a range, and a graph — which is why the next idea can talk about graphing a sequence at all. The only unusual feature is that the domain is restricted to consecutive integers.
Section
Section 2
Concept
To find a rule, rewrite each term so that its position number is visible inside it. What surrounds that number is the rule.
\[ -1,-8,-27,-64 = (-1)^3,(-2)^3,(-3)^3,(-4)^3 \]
A few terms never determine a unique rule. The sequence 2, 4, 8 fits both 2 to the n and n squared minus n plus 2, so the answer is a rule rather than the rule.
Figure (svg): Two sequences with their terms rewritten to reveal the pattern
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 795-795 — Write rules for sequences
Picture it
Example 2, both parts.
Figure (svg): Two sequences with their terms rewritten to reveal the pattern
In the first the position appears as the number being cubed; in the second it appears twice, as n minus 1 times n.
Worked example
Example 2, both parts.
\[ \text{Find the next term and a rule for } -1,-8,-27,-64,\dots \text{ and } 0,2,6,12,\dots \]
First: rewrite the terms
Why: Each is a cube of a negative integer.
\[ (-1) ^{3}, (-2) ^{3}, (-3) ^{3}, (-4) ^{3} \]
First: the next term and the rule
Why: Negative 5 cubed, and the pattern in general.
\[ -125; a _{n} = (-n) ^{3} \]
Second: rewrite the terms
Why: Each is a product of consecutive integers.
\[ 0(1), 1(2), 2(3), 3(4) \]
Second: the next term and the rule
Why: Four times 5, and the pattern in general.
\[ 20; f(n) = (n - 1) n \]
Figure (svg): Two sequences with their terms rewritten to reveal the pattern
\[ a_n = (-n)^3; \qquad f(n) = (n-1)n \]
Verify: test each rule at n equal to 3
Why: The first gives negative 3 cubed, which is negative 27 — the third term. The second gives 2 times 3, which is 6 — also the third term. Testing a rule at a middle position rather than the first is the stronger check, since many wrong rules still happen to fit n equal to 1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 795-795
Fill the middle
Example 2b.
Fill in the blanks
0, 2, 6, 12 = 0(1), 1(2), 2(3), 3(4) \;\Longrightarrow\; f(n) = (n-1)n
Why: The first factor is always one less than the position, so it is n minus 1 while the second is n. Writing each term as a product is what makes that visible.
Worked example
Guided Practice 4.
\[ \text{For } 3, 8, 15, 24, \dots \text{ find the next term and a rule.} \]
Look for a pattern
Why: Each term is one less than a perfect square.
\[ 4 - 1, 9 - 1, 16 - 1, 25 - 1 \]
Express the square in terms of n
Why: The squares are of n plus 1.
\[ (n + 1) ^{2} - 1 \]
Expand
Why: N squared plus 2n plus 1, minus 1.
\[ a _{n} = n ^{2} + 2 n \]
Find the next term
Why: Twenty-five plus 10.
\[ 35 \]
Figure (svg): The solution to Worked example one more rule shown as a ladder of expressions, one row per algebraic move
\[ a_n = n^2+2n; \quad a_5 = 35 \]
Verify: check the two forms agree
Why: N squared plus 2n factors as n times n plus 2, and at n equal to 4 that is 4 times 6, or 24 — the fourth term. The factored form n(n+2) and the expanded form describe the same rule, and either is a legitimate answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 795-795
Error analysis
A student is given the terms 2, 4, 8 and asked for the rule.
Annotate
On: \( a_n = 2^n, \text{ the only possibility} \)
The book prints this warning directly. Give a rule that fits and state the pattern you assumed, rather than claiming there is only one.
Matching
Rewrite the terms first.
Match the pairs
Why: All four are built from powers or products of the position number. The third is the fourth shifted: 3, 8, 15, 24 are the squares of 2, 3, 4, 5 each reduced by one.
Sorting
Look at how consecutive terms relate.
Sort into buckets
Sort each sequence.
The first two buckets are the arithmetic and geometric sequences of Lessons 12.2 and 12.3, and the third is everything else.
Prediction
Commit before reasoning.
Predict first
How many terms of a sequence are needed before its rule is uniquely determined?
Correct: No finite number; a rule can always be found to fit any finite list and then diverge.
\[ 2,4,8: \; 2^n \text{ gives } 16; \quad n^2-n+2 \text{ gives } 14 \]
Why: Given any n terms, a polynomial of degree n minus 1 can be found passing through all of them, and it can then be modified to take any value at the next position. So the question is not which rule is correct but which pattern is intended. That is why the book asks for a rule rather than the rule, and why stating the pattern you assumed matters as much as the formula.
Section
Section 3
Concept
To graph a sequence, put the position numbers on the horizontal axis and the terms on the vertical. Plot one point per position and leave them unjoined, since the sequence has no value between positions.
\[ (1,1), (2,4), (3,9), \dots, (7,49) \]
The points may follow a familiar curve, but drawing that curve would claim values the sequence does not have.
Figure (svg): A sequence graphed as isolated points rather than a curve
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 795-795 — Graphing sequences
Picture it
Example 3: apples per layer in a square pyramid.
Figure (svg): A sequence graphed as isolated points rather than a curve
The points lie on a parabola but the sequence is defined only at whole layers, so the curve is not drawn.
Worked example
Example 3.
\[ \text{Apples are stacked in a square pyramid of } 7 \text{ layers. Write a rule and graph it.} \]
Count the first few layers
Why: One, then a 2 by 2 square, then a 3 by 3.
\[ 1, 4, 9 \]
Recognise the pattern
Why: Each layer is a square of its layer number.
\[ a _{n} = n ^{2} \]
List the points
Why: Position on the horizontal axis, apples on the vertical.
\[ (1, 1)\text{ through } (7, 49) \]
Plot without joining
Why: The sequence has no value at layer 2.5.
Figure (svg): A sequence graphed as isolated points rather than a curve
\[ a_n = n^2 \]
Verify: check the seventh layer
Why: Seven squared is 49, and a 7 by 7 square really does hold 49 apples. The rule was read from a physical arrangement, so it can be checked against one — the bottom layer is a square with 7 apples on each side.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 795-795
Fill the middle
Guided Practice 5.
Fill in the blanks
a_9 = 9^2 = 81
Why: The ninth layer is a 9 by 9 square, holding 81 apples. The rule reaches any position directly, without listing the layers in between.
Worked example
Guided Practice 5.
\[ \text{How many apples are in the ninth layer of such a pyramid?} \]
Use the rule
Why: The layer number squared.
\[ a _{9} = 9 ^{2} \]
Compute
Why: Nine times 9.
\[ 81 \]
Check against the pattern
Why: A 9 by 9 square.
\[ 81\text{ apples} \]
Note the growth
Why: The layers grow quadratically, not linearly.
Figure (svg): The solution to Worked example extend the stack shown as a ladder of expressions, one row per algebraic move
\[ a_9 = 81 \]
Verify: compare successive layers
Why: The layers hold 1, 4, 9, 16, 25, 36, 49, 64, 81, so the differences are 3, 5, 7, 9, 11, 13, 15, 17 — the odd numbers. Constant SECOND differences of 2 confirm a quadratic rule, which is a useful pattern check on any squared sequence.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 795-795
Trap
\[ (1,1), (2,4), (3,9), \dots \]
Draw a smooth parabola through them
Why: The points follow a familiar curve, so the curve is drawn.
\[ y = x^2 \text{ for all } x \quad \text{(wrong)} \]
The curve claims a value at x equal to 2.5, but there is no layer between the second and the third.
\[ \text{plot seven points and stop} \]
Leave the points unjoined
Why: The domain is the whole numbers 1 to 7 and nothing between.
\[ \text{domain: } n = 1,2,\dots,7 \]
The book prints exactly this caution. A sequence's graph is a set of dots, however tempting the curve through them looks.
Sorting
What does a sequence's graph consist of?
Sort into buckets
Sort each statement about graphing a sequence.
The two incorrect statements are the same mistake stated twice: drawing the curve is what creates the imaginary value at 2.5.
Comparison
Fill the blanks. The same rule, different domains.
Comparison matrix
| Question | f(x) = x^2 | a_n = n^2 |
|---|---|---|
| Domain | all real numbers | the positive integers |
| Value at 2.5 | 6.25 | undefined; there is no position 2.5 |
| Graph | a continuous parabola | isolated points |
| Value at 3 | 9 | 9 |
The two agree wherever both are defined, which is exactly why the points lie on the parabola — and why drawing it anyway is still wrong.
Prediction
Commit before reasoning.
Predict first
The layer counts 1, 4, 9, 16, 25 have differences 3, 5, 7, 9. What do those differences of the differences tell you?
Correct: They are all 2, which signals a quadratic rule.
\[ 3,5,7,9 \;\to\; 2,2,2 \;\Longrightarrow\; \text{quadratic} \]
Why: Constant first differences signal a linear rule and constant second differences a quadratic one, exactly as in Lesson 11.5's shape reading. Here the first differences are the odd numbers and the second differences are all 2, confirming a degree-two rule before the formula is even guessed. A third-degree rule would show constant third differences instead.
Section
Section 4
Concept
Adding a sequence's terms gives a series. Summation notation records the rule for a term, where the index starts and where it stops, all under a sigma.
\[ 2+4+6+8 = \sum_{i=1}^{4} 2i \]
The index letter may be anything and need not start at 1, and an upper limit of infinity signals an infinite series.
Figure (svg): The parts of summation notation labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 796-796 — Series and Summation Notation
Picture it
The anatomy of summation notation.
Figure (svg): The parts of summation notation labelled
The sigma says add, the expression says what, and the two limits say from where to where. Nothing else is needed.
Worked example
Example 4, both parts.
\[ \text{Write } 25+50+75+\dots+250 \text{ and } \tfrac{1}{2}+\tfrac{2}{3}+\tfrac{3}{4}+\tfrac{4}{5}+\dots \text{ in summation notation.} \]
First: find the term rule
Why: Each term is 25 times its position.
\[ a _{i} = 25 i \]
First: find the limits
Why: The last term 250 is 25 times 10.
\[ i\text{ from } 1\text{ to } 10 \]
Second: find the term rule
Why: The denominator is one more than the numerator.
\[ a _{i} = \frac{i}{i + 1} \]
Second: find the limits
Why: The series never stops.
\[ i\text{ from } 1\text{ to } \infty \]
Figure (svg): The parts of summation notation labelled
\[ \sum_{i=1}^{10}25i; \qquad \sum_{i=1}^{\infty}\frac{i}{i+1} \]
Verify: check the upper limit
Why: Twenty-five times 10 is 250, the stated last term — so the upper limit is 10 rather than 250. Confusing the last TERM with the last INDEX is the commonest error in writing sigma notation, and substituting the proposed limit settles it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 796-796
Fill the middle
Example 4a.
Fill in the blanks
25i = 250 \;\Longrightarrow\; i = 10
Why: The last term is 250, and 25 times 10 is 250, so the index runs to 10. Solving the term rule for the last term is how the upper limit is found.
Worked example
Guided Practice 6 to 9.
\[ \text{Write } 5+10+\dots+100; \; \tfrac{1}{2}+\tfrac{4}{5}+\tfrac{9}{10}+\tfrac{16}{17}+\dots; \; 6+36+216+\dots; \; 5+6+7+\dots+12. \]
First: multiples of 5 up to 100
Why: One hundred is 5 times 20.
\[ \sum\text{ of } 5 i, i\text{ from } 1\text{ to } 20 \]
Second: squares over squares plus one
Why: One over 2, then 4 over 5, then 9 over 10.
\[ \sum\text{ of } i ^{2} / (i ^{2} + 1)\text{ to } \infty \]
Third: powers of 6
Why: Six, 6 squared, 6 cubed, 6 to the fourth.
\[ \sum\text{ of } 6 ^{i}\text{ to } \infty \]
Fourth: consecutive integers from 5
Why: The index itself is the term.
\[ \sum\text{ of } i, i\text{ from } 5\text{ to } 12 \]
Figure (svg): The solution to Worked example four more series shown as a ladder of expressions, one row per algebraic move
\[ \sum_{i=1}^{20}5i; \; \sum_{i=1}^{\infty}\tfrac{i^2}{i^2+1}; \; \sum_{i=1}^{\infty}6^i; \; \sum_{i=5}^{12}i \]
Verify: check the last one's limits
Why: Starting the index at 5 rather than 1 lets the term rule be simply i. The alternative, starting at 1 with a term rule of i plus 4, gives the same series — so more than one correct sigma expression exists for most series.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 796-796
Error analysis
A student writes 25 plus 50 plus 75 up to 250 in summation notation.
Annotate
On: \( \sum_{i=1}^{250}25i \)
Substituting the proposed upper limit into the term rule is the check: 25 times 10 is 250, and 25 times 250 is 6250.
Matching
Find the rule, then the limits.
Match the pairs
Why: The third has the index as an exponent rather than a multiplier, which is the difference between an arithmetic and a geometric series. The fourth starts its index at 5, which is allowed and often the simplest choice.
Sorting
Four pieces of one symbol.
Sort into buckets
Sort each element of summation notation.
Reading a sigma expression aloud in that order — add this, from here, to there — is the fastest way to expand one correctly.
Prediction
Commit before reasoning.
Predict first
The series 5 plus 6 plus 7 up to 12 can be written as the sum of i from 5 to 12. Is that the only way?
Correct: No — the sum of i plus 4 from 1 to 8 gives the same series.
\[ \sum_{i=5}^{12}i = \sum_{i=1}^{8}(i+4) \]
Why: Shifting the starting index and adjusting the term rule to compensate leaves the series unchanged: i from 5 to 12 and i plus 4 from 1 to 8 both produce 5, 6, 7 up to 12. Either is correct, and the one with the simpler term rule is usually preferred. This flexibility becomes useful in Lesson 12.3, where shifting an index simplifies a geometric sum.
Section
Section 5
Concept
A short series can be evaluated by writing out the terms and adding. Three special series have closed formulas that replace the addition entirely.
\[ \sum_{i=1}^{n}i = \frac{n(n+1)}{2}; \quad \sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6} \]
The formulas apply only when the index runs from 1, so a series starting elsewhere must be adjusted before they can be used.
Figure (svg): The three special sum formulas with a worked application
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 797-797 — Formulas for Special Series
Picture it
The sums of ones, of integers and of squares.
Figure (svg): The three special sum formulas with a worked application
The apple pyramid's total comes out of the third in one line: 7 times 8 times 15, all over 6, which is 140.
Worked example
Example 5.
\[ \text{Find } \sum_{k=4}^{8}(3+k^2). \]
Substitute each index value
Why: K equal to 4, 5, 6, 7 and 8.
Evaluate the terms
Why: Three plus 16, three plus 25, and so on.
\[ 19, 28, 39, 52, 67 \]
Add them
Why: Nineteen plus 28 plus 39 plus 52 plus 67.
\[ 205 \]
Note the index
Why: It runs from 4, not from 1.
Figure (svg): The solution to Worked example expand a series shown as a ladder of expressions, one row per algebraic move
\[ 19+28+39+52+67 = 205 \]
Verify: check by splitting the sum
Why: The series is five 3s plus the squares of 4 through 8, which is 15 plus 16 plus 25 plus 36 plus 49 plus 64, or 15 plus 190, giving 205. Splitting a sum into simpler pieces is both a check and often a shortcut.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 797-797
Fill the middle
Example 6.
Fill in the blanks
\sum_140^___i^2 = \frac______ = \frac______ = ___
Why: Eight hundred forty over 6 is 140, the total apples in the stack. One evaluation replaced seven squarings and six additions.
Worked example
Example 6 and Guided Practice 14.
\[ \text{Find the total apples in a } 7 \text{-layer pyramid, and in a } 9 \text{-layer one.} \]
Write the series
Why: The layers hold i squared apples for i from 1 to 7.
\[ \sum\text{ of } i ^{2}, 1\text{ to } 7 \]
Apply the formula
Why: Seven times 8 times 15, over 6.
\[ \frac{840}{6} = 140 \]
Repeat for nine layers
Why: Nine times 10 times 19, over 6.
\[ \frac{1710}{6} \]
Compute
Why: One thousand seven hundred ten over 6.
\[ 285 \]
Figure (svg): The three special sum formulas with a worked application
\[ 140; \qquad 285 \]
Verify: check the seven-layer total by adding
Why: One plus 4 plus 9 plus 16 plus 25 plus 36 plus 49 is 140, matching the formula. Two extra layers took the total from 140 to 285, more than doubling it — because the new layers hold 64 and 81, the two largest in the stack.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 797-797
Trap
\[ \sum_{k=3}^{7}(k^2-1) \]
Use the sum of squares formula with n equal to 7
Why: The upper limit is read as the formula's n.
\[ \frac{7(8)(15)}{6}-1 = 139 \quad \text{(wrong)} \]
The formula sums from 1, but this series starts at 3 — and the minus 1 appears in every one of the five terms, not once.
\[ (9-1)+(16-1)+(25-1)+(36-1)+(49-1) = 130 \]
Expand, or subtract the missing initial terms
Why: The formulas require the index to start at 1.
\[ \sum_{k=1}^{7}k^2-\sum_{k=1}^{2}k^2-5 = 140-5-5 = 130 \]
Both routes give 130. Subtracting the unwanted early terms is the general way to use a formula on a shifted index.
Matching
Expand or apply a formula.
Match the pairs
Why: The third looks strange until you notice that adding 1 thirty-four times gives 34 — the term rule does not mention the index at all. The first factors out the 8 and uses the sum of the first five integers, 15.
Ranking
Smallest first.
Put in order
Why: The values are 21, 34, 120, 140 and 285. The last two show how fast a sum of squares grows: two extra terms nearly doubled the total, because the terms themselves are growing quadratically.
Prediction
Commit before reasoning.
Predict first
Why does 1 plus 2 up to n equal n times n plus 1, all over 2?
Correct: Because pairing the first with the last, the second with the second-last and so on gives n over 2 pairs each summing to n plus 1.
\[ \frac{n}{2}\text{ pairs} \times (n+1) = \frac{n(n+1)}{2} \]
Why: For n equal to 100 the pairs are 1 plus 100, 2 plus 99, 3 plus 98 and so on — fifty pairs each totalling 101, giving 5050. The argument works for odd n too, with the middle term left over at half the pair value. The pairing trick is the classic proof, and it is the same idea that will give the arithmetic series formula in Lesson 12.2.
Comparison
Fill the blanks. One is a list, the other a total.
Comparison matrix
| Question | Sequence | Series |
|---|---|---|
| What it is | a list of terms | the terms added together |
| Example | 2, 4, 6, 8 | 2 + 4 + 6 + 8 = 20 |
| Written as | a rule for a_n | summation notation |
| The answer is | a list | a single number |
The rule inside a sigma is the sequence's rule, so writing a series always begins by finding the sequence behind it.
Pattern
Terms, rule, then sum.
The upper limit counts index values, not term values. Substitute it into the term rule to check.
OpenStax Algebra and Trigonometry 2e, §13.1 Sequences and Their Notations §13.1
Check
Substitute the positions.
Check your understanding
What are the first three terms of f(n) = (-3)^(n-1)?
Answer: A
Why: At n = 1 the exponent is 0, giving 1, then -3 and 9.
Check
The limit counts index values.
Check your understanding
How is 25 + 50 + 75 + ... + 250 written in summation notation?
Answer: A
Why: The last term 250 is 25 times 10, so the index runs to 10.
Check
Use the squares formula.
Check your understanding
What is the sum of i squared for i from 1 to 7?
Answer: A
Why: Seven times 8 times 15, all over 6, gives 840 over 6.
Real world
A theatre has 20 rows. The front row holds 18 seats and each row behind holds two more than the one in front.
Discussion prompt
Write a rule for the seats in row n, and find the total number of seats.
Hint: Write the total as a sum you can split.
Answer:
\[ a_n = 18+2(n-1) = 2n+16 \]
\[ \text{total} = \sum_{n=1}^{20}(2n+16) = 2\sum_{n=1}^{20}n + \sum_{n=1}^{20}16 \]
\[ = 2\left(\frac{20\cdot 21}{2}\right)+20(16) = 420+320 = 740 \]
The theatre holds 740 seats, found without adding twenty numbers.
The split into two sums is the technique worth keeping: a sigma distributes over addition and constants come out front, so a complicated term rule breaks into pieces the special formulas can handle. The same three formulas cover an enormous range of problems once you can split a sum this way — and Lesson 12.2 will give a single formula for exactly this kind of series, where each term exceeds the last by a fixed amount.
Commit first
Answer, then rate your confidence honestly.
Predict first
The sequence 2, 4, 8 obviously continues 16. Is the rule necessarily 2 to the n?
Correct: No — n squared minus n plus 2 also gives 2, 4, 8, and then 14.
\[ 2^4 = 16; \quad 4^2-4+2 = 14 \]
Why: Both rules reproduce the three given terms exactly and disagree from the fourth onward. No finite list of terms determines a rule, because a polynomial can always be built through any finite set of points and then bent afterwards. The book prints this warning beside Example 2. The right response is not to give up but to state a rule that fits together with the pattern you assumed, so a reader can see what was taken for granted.
Explain it
They can evaluate a function and have never seen sigma notation.
Discussion prompt
In four sentences or fewer, explain what a sigma expression tells you to do.
Hint: Read the parts in order.
Answer:
The big sigma means add things up. Underneath it says where to start, above it says where to stop, and beside it is the recipe for each term.
So the sum of 2i from i equals 1 to 4 means: work out 2i for i equal to 1, 2, 3 and 4, then add the results. That gives 2 plus 4 plus 6 plus 8, which is 20.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For off-by-one errors, always test the rule at n equal to 1 first. For finding rules, rewrite each term until the position number appears inside it. For upper limits, substitute your proposed limit into the term rule and check it gives the last term. For formulas, check that the index starts at 1 and the terms are ones, integers or squares.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a sequences page. Top left: draw the domain-to-range arrow diagram for a rule of your own and list its first six terms. Top right: take three sequences of terms and rewrite each term until the position number is visible, then write a rule for each, noting beside one of them a second rule that also fits the given terms. Middle: graph the apple pyramid sequence as seven isolated points and write one sentence on why the parabola through them is not drawn. Bottom left: draw a large sigma and label all four of its parts, then expand two sigma expressions in full. Bottom right: write the three special formulas, use each once, and show how a sum starting at index 3 can be handled by subtracting the first two terms.
If any of your sigma expressions has the last term as its upper limit, recheck it: the limit counts positions, not values.
Recap
Five things, and a compact notation for all of them.
| If you see | Then |
|---|---|
| A rule with n in it | Substitute 1, 2, 3 to get the terms |
| A handful of terms | Rewrite them until n appears inside each |
| A sequence to graph | Plot isolated points; never join them |
| A series to write | Find the term rule, then both limits |
| A sigma with an upper limit | It counts index values, not term values |
| A sum of integers or squares from 1 | Use the closed formula |
Lesson 12.2 takes the sequences whose terms differ by a constant and gives closed formulas for both their terms and their sums.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.1 Define and Use Sequences and Series §12.1, pp. 794-801 — everything on these slides traces back here
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