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AMC 8
Algebra · Chapter 16

Telescoping

Write out the first few and last few terms of a giant sum or product, watch the middle cancel away, and keep what is left.

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16.10%

Telescoping

In this section you will be able to see a long sum fold up like a telescope, so only a few pieces are left.

A pirate’s telescope is long when it’s open. Push the end and each tube slides into the next one. Soon it’s tiny! Some sums do the same thing.

Each fraction 1/(k(k+1)) splits into two pieces: 1/k − 1/(k+1). The minus piece of one term is the plus piece of the next term. They cancel!

Look:after all the cancelling, only the very first piece and the very last piece are left.

Exploreplay first, formulas later

Sum of 1/(k(k+1)) for k = 1 to 4: each −1/(k+1) cancels the next +1/(k+1), leaving 1/1 − 1/5 = 4/5

last k4
1/(1·2)(11 − 12)
+1/(2·3)(12 − 13)
+1/(3·4)(13 − 14)
+1/(4·5)(14 − 15)
0 of 3 cancellations

0 of 3 cancellations done

Checkpoint for section 16.1

Get 2 of 3 right to master this section. One try each, no hints.
1

Find 11⋅2+12⋅3+13⋅4\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4}.

2

Find 21⋅32⋅43⋅54\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4}.

3

After all the cancelling in a telescoping sum, what is left?

16.20%

Telescoping Basics

In this section you will be able to spot a telescoping problem and cancel the middle by writing the first few and last few terms.

What is 1 − 2 + 3 − 4 + 5 − 6 + … all the way to 1000? Adding a thousand numbers by hand is no fun. There must be a shortcut!

You never write out every term. Write the first few and the last few. Look for pieces that cancel, or blocks that are all worth the same.

Look:if a problem looks impossible to do by hand, telescoping is probably the trick.

Exploreplay first, formulas later

20 terms: 10 blocks of 2, each worth -1; total -10

block size2
( 1 − 2 )= ?( 3 − 4 )= ?( 5 − 6 )= ?⋯( 17 − 18 )= ?( 19 − 20 )= ?
0 of 5 blocks added up

0 of 5 blocks added up

Name it

Concept 16.2.1 · Telescoping

Expand the first few and last few terms, and cancel out any terms you see.

Remark 16.2.2

Generally, whenever you have long expressions that seem to be hard or impossible to compute manually, telescoping is probably at play.

Try it

Problem 16.6.4

AMC 8

Find the value of the expression 100−98+96−94+92−90+⋯+8−6+4−2100 - 98 + 96 - 94 + 92 - 90 + \cdots + 8 - 6 + 4 - 2.

Problem 16.6.1

AMC 8

What is the value of 4⋅(−1+2−3+4−5+6−7+⋯+1000)4 \cdot (-1 + 2 - 3 + 4 - 5 + 6 - 7 + \cdots + 1000)?

Checkpoint for section 16.2

Get 2 of 3 right to master this section. One try each, no hints.
1

Find 1−2+3−4+⋯+19−201 - 2 + 3 - 4 + \cdots + 19 - 20.

2

Find 10−9+8−7+6−5+4−3+2−110 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1.

3

Find (12−13)+(13−14)+⋯+(199−1100)\left(\frac12 - \frac13\right) + \left(\frac13 - \frac14\right) + \cdots + \left(\frac{1}{99} - \frac{1}{100}\right).

16.30%

Telescoping Sums

In this section you will be able to add long lists of fractions and signed squares by cancelling or grouping.

Can you add 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 in your head? It sounds hard. But these fractions are secretly differences!

The key fact: 1/(k(k+1)) = 1/k − 1/(k+1). Split every term, and the middle pieces cancel in pairs.

Long lists with plus and minus signs work too. Find a block size where every block has the same value. Then count the blocks.

Exploreplay first, formulas later

Sum of 1/(k(k+1)) for k = 1 to 9: each −1/(k+1) cancels the next +1/(k+1), leaving 1/1 − 1/10 = 9/10

last k9
1/(1·2)(11 − 12)
+1/(2·3)(12 − 13)
+1/(3·4)(13 − 14)
+⋯
+1/(8·9)(18 − 19)
+1/(9·10)(19 − 110)
0 of 5 cancellations

0 of 5 cancellations done

Watch itpredict, then reveal

Example 16.1

EMCC

Evaluate (12−32+52−72+92−⋯+20092)−(22−42+62−82+102−⋯+20102)(1^2 - 3^2 + 5^2 - 7^2 + 9^2 - \cdots + 2009^2) - (2^2 - 4^2 + 6^2 - 8^2 + 10^2 - \cdots + 2010^2)

2010 terms: 1005 blocks of 2 that are not all equal; total -2011

block size2
( 1² − 2² )= ?( −3² + 4² )= ?( 5² − 6² )= ?⋯( −2007² + 2008² )= ?( 2009² − 2010² )= ?
0 of 5 blocks added up

0 of 5 blocks added up

  1. 1

    Open both brackets and put the terms in order 12,22,32,…1^2, 2^2, 3^2, \ldots What signs do you get?

    Predict first:
open on YouTube

Example 16.2

Find 16+112+120+130+142+156+172\frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72}

Sum of 1/(k(k+1)) for k = 2 to 8: each −1/(k+1) cancels the next +1/(k+1), leaving 1/2 − 1/9 = 7/18

1/(2·3)(12 − 13)
+1/(3·4)(13 − 14)
+1/(4·5)(14 − 15)
+1/(5·6)(15 − 16)
+1/(6·7)(16 − 17)
+1/(7·8)(17 − 18)
+1/(8·9)(18 − 19)
0 of 6 cancellations

0 of 6 cancellations done

  1. 1

    Look at the denominators: 6=2⋅36 = 2 \cdot 3, 12=3⋅412 = 3 \cdot 4, 20=4⋅520 = 4 \cdot 5. The last one is 72=k(k+1)72 = k(k+1). What is kk?

    Predict first:
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Try it

Problem 16.6.10(additional)

Evaluate 12×3+13×4+⋯+110×11\frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{10 \times 11}

Contest cornerreal contest problems

Problem 16.6.6(additional)

AMC 8

What is the value of 1+3+5+⋯+2017+2019−2−4−6−⋯−2016−20181 + 3 + 5 + \cdots + 2017 + 2019 - 2 - 4 - 6 - \cdots - 2016 - 2018?

Problem 16.6.9(additional)

MATHCOUNTS

If n=102020−102019+102018−102017+⋯+102−101n = 10^{2020} - 10^{2019} + 10^{2018} - 10^{2017} + \cdots + 10^2 - 10^1 what is the sum of digits of the integer nn?

Problem 16.6.11(additional)

Omega Learn

Evaluate 1−2−3+4+5+6−7−8−9−10⋯+90+911 - 2 - 3 + 4 + 5 + 6 - 7 - 8 - 9 - 10 \cdots + 90 + 91 (1 positive consecutive value, 2 negative consecutive values, 3 positive consecutive values, …)

Checkpoint for section 16.3

Get 2 of 3 right to master this section. One try each, no hints.
1

Find 11⋅2+12⋅3+⋯+19⋅10\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{9 \cdot 10}.

2

Find 112+120+130\frac{1}{12} + \frac{1}{20} + \frac{1}{30}.

3

Find 22−12+42−32+62−522^2 - 1^2 + 4^2 - 3^2 + 6^2 - 5^2.

16.40%

Telescoping Products

In this section you will be able to multiply long chains of fractions where each top cancels a bottom.

What is 3/2 × 4/3 × 5/4 × … × 2006/2005? That’s over two thousand fractions! Look closely at the 3s and the 4s.

Products telescope too. When the top of one fraction matches the bottom of another, they cancel. Only a few tops and bottoms survive.

Look:sometimes a top cancels a bottom two fractions away, not the very next one. Check what is left at both ends.

Exploreplay first, formulas later

Product of 4 fractions for k = 1 to 4: after cancelling, 5 / 1 = 5

last k4
21
×32
×43
×54
0 of 3 cancellations

0 of 3 cancellations done

Watch itpredict, then reveal

Example 16.3

Omega Learn

What is the value of 34×46×58×⋯×1018\frac{3}{4} \times \frac{4}{6} \times \frac{5}{8} \times \cdots \times \frac{10}{18}? (For each fraction, the denominator is twice the numerator minus 2)

Product of 8 fractions for k = 3 to 10: after cancelling, 10 / 2^9 = 5/256

32·2
×42·3
×52·4
×62·5
×72·6
×82·7
×92·8
×102·9
0 of 7 cancellations

0 of 7 cancellations done

  1. 1

    The denominator is 2n−2=2(n−1)2n - 2 = 2(n-1). So 34=32⋅2\frac{3}{4} = \frac{3}{2 \cdot 2} and 46=42⋅3\frac{4}{6} = \frac{4}{2 \cdot 3}. How many fractions are there, from 34\frac{3}{4} to 1018\frac{10}{18}?

    Predict first:
open on YouTube

Try it

Problem 16.6.3

AMC 8

What is the value of the product (1+11)⋅(1+12)⋅(1+13)⋅(1+14)⋅(1+15)⋅(1+16)\left(1+\frac{1}{1}\right) \cdot \left(1+\frac{1}{2}\right) \cdot \left(1+\frac{1}{3}\right) \cdot \left(1+\frac{1}{4}\right) \cdot \left(1+\frac{1}{5}\right) \cdot \left(1+\frac{1}{6}\right)?

Problem 16.6.7(additional)

AMC 8

What is the product of 32×43×54×⋯×20062005\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \cdots \times \frac{2006}{2005}?

Contest cornerreal contest problems

Problem 16.6.2

AMC 8

What is the value of the expression 1⋅2⋅3⋅4⋅5⋅6⋅7⋅81+2+3+4+5+6+7+8\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}?

Problem 16.6.5

AMC 8

What is the value of the product (1⋅32⋅2)(2⋅43⋅3)(3⋅54⋅4)⋯\left(\frac{1 \cdot 3}{2 \cdot 2}\right)\left(\frac{2 \cdot 4}{3 \cdot 3}\right)\left(\frac{3 \cdot 5}{4 \cdot 4}\right) \cdots (97⋅9998⋅98)(98⋅10099⋅99)\left(\frac{97 \cdot 99}{98 \cdot 98}\right)\left(\frac{98 \cdot 100}{99 \cdot 99}\right)?

Problem 16.6.8(additional)

AMC 8

What is the value of 13⋅24⋅35⋯1820⋅1921⋅2022\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdots \frac{18}{20} \cdot \frac{19}{21} \cdot \frac{20}{22}?

Checkpoint for section 16.4

Get 2 of 3 right to master this section. One try each, no hints.
1

Find (1−12)(1−13)\left(1 - \frac12\right)\left(1 - \frac13\right) (1−14)⋯(1−110)\left(1 - \frac14\right) \cdots \left(1 - \frac{1}{10}\right).

2

Find 43⋅54⋅65⋯3029\frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdots \frac{30}{29}.

3

Find 13⋅24⋅35⋅46⋅57\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \frac{4}{6} \cdot \frac{5}{7}.

16.50%

Telescoping Equation

In this section you will be able to subtract neighbouring equations so the shared unknown cancels.

You have 2021 equations and 2022 unknowns. That sounds hopeless! But each equation shares one unknown with the next one.

Subtract one equation from the next. The shared unknown cancels, just like in a telescoping sum. You get x3−x1=1x_3 - x_1 = 1, x4−x2=1x_4 - x_2 = 1, and so on.

So every second unknown goes up by exactly 1. Write everything in terms of x1x_1, then use the extra clue to find x1x_1.

Exploreplay first, formulas later

5 equations x_k + x_(k+1) = k: subtracting neighbours cancels the shared x and gives x_(k+2) − x_k = 1, so x_2022 = 1011 − x₁

equations5
  1. x1 + x2 = 1
  2. x2 + x3 = 2
  3. x3 + x4 = 3
  4. x4 + x5 = 4
  5. x5 + x6 = 5
  6. ⋮
  1. Subtract each equation from the next one.
0 of 4 subtractions

0 of 4 subtractions done

Watch itpredict, then reveal

Example 16.4

BmMT

Suppose x1,x2,…,x2022x_1, x_2, \ldots, x_{2022} is a sequence of real numbers such that:

x1+x2=1x2+x3=2⋮x2021+x2022=2021\begin{aligned} x_1 + x_2 &= 1 \\ x_2 + x_3 &= 2 \\ & \vdots \\ x_{2021} + x_{2022} &= 2021 \end{aligned}

If x1+x499+x999+x1501=222x_1 + x_{499} + x_{999} + x_{1501} = 222, then what is the value of x2022x_{2022}?

5 equations x_k + x_(k+1) = k: subtracting neighbours cancels the shared x and gives x_(k+2) − x_k = 1, so x_2022 = 1011 − x₁

equations5
  1. x1 + x2 = 1
  2. x2 + x3 = 2
  3. x3 + x4 = 3
  4. x4 + x5 = 4
  5. x5 + x6 = 5
  6. ⋮
  1. Subtract each equation from the next one.
0 of 4 subtractions

0 of 4 subtractions done

  1. 1

    Subtract the first equation from the second: (x2+x3)−(x1+x2)(x_2 + x_3) - (x_1 + x_2). What is x3−x1x_3 - x_1?

    Predict first:
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Checkpoint for section 16.5

Get 2 of 3 right to master this section. One try each, no hints.
1

If x1+x2=1x_1 + x_2 = 1 and x2+x3=2x_2 + x_3 = 2, what is x3−x1x_3 - x_1?

2

Suppose xk+xk+1=kx_k + x_{k+1} = k for every kk, and x1=5x_1 = 5. What is x7x_7?

3

Same chain xk+xk+1=kx_k + x_{k+1} = k with x1=5x_1 = 5. What is x6x_6?

16.6

Practice Problems

In this section you will be able to mix telescoping sums and products on a tough challenge.

Sohil’s messy handwriting turned a product into a sum! Can you clean up after him? Break the expression into blocks, and telescope each block.

Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 16.6.12(additional)

Omega Learn

Let the sequence ana_n be defined as an=31+2+3+⋯+na_n = 3^{1+2+3+\cdots+n} for all positive integers nn. Sohil is trying to evaluate the following sequence: 12×23×34×…\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots. Due to his bad handwriting, in the expression above, he accidentally writes each multiplication sign before all fractions of the form anan+1\frac{a_n}{a_n+1} (for example, 3131+1\frac{3^1}{3^1+1}, 3(1+2)3(1+2)+1\frac{3^{(1+2)}}{3^{(1+2)}+1}, etc.) with a plus sign. What is the value of the new expression?