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Telescoping Sums: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra I. You can skip it.

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Problem 1 of 10

Problem 1 The closed record

Three real readings satisfy $b_1-b_2=7$ and $b_2-b_3=-2$ . Find $b_3-b_1$ .
1. Hint 1
  Adding the recorded differences cancels the middle reading.
2. Hint 2
  Reverse the direction of the resulting difference.
Answer

$b_3-b_1=-5$ .

Full solution

Adding the two equations cancels $b_2$ , giving
$b_1-b_3=7+(-2).$
Thus $b_1-b_3=5$ .
Reversing the difference changes its sign, so
$b_3-b_1=-5.$
The three directed changes total $7-2-5=0$ , as a closed record should.

Answer

$b_3-b_1=-5$ .

Key idea

Differences around a closed chain add to zero because every reading appears with both signs.
Problem 1: Mark complete
Problem 2 The quadratic denominators

Evaluate
$\sum_{k=3}^{14}\frac{1}{k^2+k}$
exactly, as a fraction in lowest terms.
1. Hint 1
  Factor the denominator before hunting for a split.
2. Hint 2
  Each fraction becomes $1/k-1/(k+1)$ , so line the brackets up.
3. Hint 3
  Only the first positive fraction and the last negative fraction lack a partner.
Answer

$\frac{4}{15}$ .

Full solution

Factoring gives $k^2+k=k(k+1)$ , and every denominator is nonzero because $k\ge3$ .
Combining $1/k-1/(k+1)$ over the common denominator rebuilds that term, so the split
$\frac{1}{k(k+1)}=\frac1k-\frac{1}{k+1}$
is exact.
Each interior fraction is subtracted in one bracket and added back in the next, so it cancels.
The survivors are the leading $1/3$ and the final $-1/15$ , giving
$\frac13-\frac1{15}=\frac{4}{15}.$
As a check, the first two terms are $1/12+1/20=2/15$ , matching the survivors $1/3-1/5$ of those two brackets.

Answer

$\frac{4}{15}$ .

Key idea

Factoring a quadratic denominator can reveal the difference of reciprocals that makes a sum telescope.
Problem 2: Mark complete
Problem 3 The boundary reading

A record satisfies $\sum_{k=3}^{7}(b_k-b_{k+2})=5$ . It also gives $b_3=8$ , $b_4=-1$ , and $b_8=4$ . Find $b_9$ .
1. Hint 1
  List the positive and negative subscripts in the sum.
2. Hint 2
  The interior readings cancel two positions apart.
3. Hint 3
  Use the four boundary readings to form one equation.
Answer

$b_9=-2$ .

Full solution

The positive readings run from $b_3$ through $b_7$ , while the negative readings run from $b_5$ through $b_9$ .
Canceling the interior readings gives
$b_3+b_4-b_8-b_9=5.$
Substitution gives
$8-1-4-b_9=5.$
Thus $b_9=-2$ .
The surviving total checks as
$8-1-4-(-2)=5.$

Answer

$b_9=-2$ .

Key idea

A gap of two leaves two boundary readings at each end, even when the interior readings are unknown.
Problem 3: Mark complete
Problem 4 The omitted radical

A report should add the terms $\sqrt{k+1}-\sqrt{k}$ for the integers $k=1$ through $k=7$ , but it omits the term with $k=4$ . Find the reported total in simplified exact form.
1. Hint 1
  Find the complete total by cancellation.
2. Hint 2
  Write out the single omitted term.
3. Hint 3
  Subtract that entire term, keeping its signs.
Answer

$2\sqrt2+1-\sqrt5$ .

Full solution

The complete total cancels to $\sqrt8-1$ .
The omitted term is $\sqrt5-2$ .
Therefore the reported total is
$R=(\sqrt8-1)-(\sqrt5-2).$
Simplifying gives
$R=2\sqrt2+1-\sqrt5.$
As a check, the two retained blocks cancel separately to $2-1$ and $\sqrt8-\sqrt5$ , giving the same result.

Answer

$2\sqrt2+1-\sqrt5$ .

Key idea

A missing term breaks a cancellation chain, so subtract it explicitly or total the retained blocks separately.
Problem 4: Mark complete
Problem 5 The painted strips

A square sheet has side length $7$ centimeters. Its side is increased by $1$ centimeter at a time until it reaches $20$ centimeters, and each increase paints the strip of material it adds. Write the painted total as a sum of differences of consecutive squares, then find that total in square centimeters and state how many strips are painted.
1. Hint 1
  Each increase adds the new square area minus the old square area.
2. Hint 2
  Chain the strips and look for each square area appearing with both signs.
3. Hint 3
  Count the stages carefully: the first strip carries the side from $7$ to $8$ .
Answer

$\sum_{s=7}^{19}\left[(s+1)^2-s^2\right]=351$ square centimeters, painted in $13$ strips.

Full solution

Raising the side from $s$ to $s+1$ adds a strip of area $(s+1)^2-s^2$ , because the larger square contains the smaller one.
The painted total is therefore
$\sum_{s=7}^{19}\left[(s+1)^2-s^2\right].$
Each $-s^2$ meets the $+s^2$ of the next strip, so every interior square area cancels.
Only the largest square area and the smallest survive, leaving
$20^2-7^2=351$
square centimeters.
The stages run from $s=7$ to $s=19$ , so $19-7+1=13$ strips are painted.
As a check, the strip areas are the odd numbers $15,17,\ldots,39$ , whose average is $27$ , and
$13\cdot27=351.$

Answer

$\sum_{s=7}^{19}\left[(s+1)^2-s^2\right]=351$ square centimeters, painted in $13$ strips.

Key idea

Differences of consecutive square areas telescope, so a long chain of added strips collapses to the final area minus the starting area.
Problem 5: Mark complete
Problem 6 The target total

Let $T_n=\sum_{k=1}^{n}1/[k(k+2)]$ for positive integers $n$ . Find the least $n$ for which $T_n>\frac23$ , and explain why no smaller index works.
1. Hint 1
  Split $1/[k(k+2)]$ into half of $1/k-1/(k+2)$ .
2. Hint 2
  Keep two positive boundary fractions and two negative ones.
3. Hint 3
  The totals increase, so compare the indices on either side of the crossing.
Answer

$n=11$ .

Full solution

Combining fractions verifies
$\frac{1}{k(k+2)}=\frac12\left(\frac1k-\frac1{k+2}\right).$
The gap of two cancels each negative fraction two terms later.
The positive survivors total $1+1/2=3/2$ .
The outer factor $1/2$ halves those survivors and the leftover tail alike, so the halved survivors give $3/4$ .
Define the remaining tail by
$E_n=\frac12\left(\frac1{n+1}+\frac1{n+2}\right).$
Then $T_n=3/4-E_n$ , so the target requires $E_n<1/12$ .
At $n=10$ , $E_{10}=23/264$ , which exceeds
$1/12=22/264.$
At $n=11$ , $E_{11}=25/312$ , which is less than
$1/12=26/312.$
Every added term is positive, so all earlier totals are smaller.
Hence $11$ is the least index.

Answer

$n=11$ .

Key idea

For positive telescoping terms, a strict target can be located by comparing the surviving tails at neighboring indices.
Problem 6: Mark complete
Problem 7 The nine differences

For a positive integer $m$ , nine differences satisfy
$\sum_{k=m}^{m+8}(\sqrt{k+1}-\sqrt{k})=1.$
Find $m$ and check it in the uncanceled boundary expression.
1. Hint 1
  The nine differences leave one square root at each end.
2. Hint 2
  Isolate one square root before squaring.
3. Hint 3
  Check the result in the boundary equation because squaring may introduce unwanted values.
Answer

$m=16$ ; $\sqrt{25}-\sqrt{16}=1$ .

Full solution

Cancellation leaves
$\sqrt{m+9}-\sqrt m=1.$
Isolating the first root gives
$\sqrt{m+9}=1+\sqrt m.$
Both sides are positive.
Squaring gives
$m+9=1+2\sqrt m+m.$
Therefore $\sqrt m=4$ , so $m=16$ .
In the original boundary expression, $\sqrt{25}-\sqrt{16}=5-4=1$ , and the indices $16$ through $24$ contain exactly nine terms.

Answer

$m=16$ ; $\sqrt{25}-\sqrt{16}=1$ .

Key idea

Telescoping can turn an unknown starting position into an equation involving just the two boundary values.
Problem 7: Mark complete
Problem 8 The raised reading

For $n\ge3$ , let $S=\sum_{k=1}^{n}(b_k-b_{k+1})$ . One reading $b_j$ , with $2\le j\le n$ , is increased by $7$ wherever it appears. A student says $S$ is unchanged. Is this correct? Explain.
1. Hint 1
  Locate both appearances of the changed reading.
2. Hint 2
  It appears negatively in one difference and positively in the next.
3. Hint 3
  Compare the changes to those two terms.
Answer

Yes; $S$ is unchanged.

Full solution

The reading $b_j$ appears in $b_{j-1}-b_j$ and in $b_j-b_{j+1}$ .
Increasing it by $7$ lowers the first difference by $7$ and raises the second by $7$ .
The net change is
$-7+7=0.$
Equivalently, the sum is $b_1-b_{n+1}$ , and neither boundary reading changed.

Answer

Yes; $S$ is unchanged.

Key idea

Changing an interior reading consistently in both neighboring differences leaves a telescoping total unchanged.
Problem 8: Mark complete
Problem 9 The logarithm chain

Decide whether $\sum_{k=1}^{\infty}\log_{10}\left(\frac{k+1}{k}\right)$ has a finite total, justifying your verdict from its partial sums.
1. Hint 1
  The logarithm of a quotient is a difference of two logarithms.
2. Hint 2
  After that rewrite, the interior logarithms cancel in pairs.
3. Hint 3
  Compare the totals at $n=9$ , at $n=99$ and at $n=999$ before deciding.
Answer

No finite total; the partial sums are $S_n=\log_{10}(n+1)$ .

Full solution

Write $S_n$ for the total of the first $n$ terms.
The quotient rule writes each term as the difference
$\log_{10}(k+1)-\log_{10}(k),$
a mirror telescope with
$b_k=\log_{10}(k).$
Every interior logarithm is added once and subtracted once, so only the two boundary values survive, giving
$S_n=\log_{10}(n+1)-\log_{10}(1).$
Since $\log_{10}(1)=0$ , the partial sums are
$S_n=\log_{10}(n+1).$
Now test whether that surviving expression settles as $n$ climbs.
For any positive integer $m$ , taking $n=10^m-1$ gives $S_n=m$ , so the partial sums pass every whole number instead of approaching one fixed value.
The infinite sum therefore has no finite total, even though each term shrinks toward zero as $(k+1)/k$ moves toward $1$ .

Answer

No finite total; the partial sums are $S_n=\log_{10}(n+1)$ .

Key idea

Terms shrinking toward zero do not by themselves give a finite infinite total; what must settle is the surviving boundary expression.
Problem 9: Mark complete
Problem 10 The student's shortcut

Let $b_k=3+\frac1{k+2}$ for positive integers $k$ . A student says $\sum_{k=1}^{\infty}(b_k-b_{k+1})=b_1$ because the interior cancels. Decide whether the claim is correct and give the exact sum.
1. Hint 1
  Write the finite partial sum first.
2. Hint 2
  Determine what the final boundary reading approaches.
3. Hint 3
  Subtract that limiting boundary value from the initial reading.
Answer

The claim is false; the sum is $\frac13$ .

Full solution

The finite sum is $b_1-b_{n+1}$ .
Here $b_1=10/3$ and $b_{n+1}=3+1/(n+3)$ , so
$S_n=\frac13-\frac1{n+3}.$
The fraction $1/(n+3)$ tends to zero, and the partial sums approach $1/3$ .
The final reading itself approaches $3$ , not zero, so discarding the entire final reading would incorrectly give $b_1$ .

Answer

The claim is false; the sum is $\frac13$ .

Key idea

When the final boundary reading settles, a telescoping infinite total subtracts that limiting value, which need not be zero.
Problem 10: Mark complete

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