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PS · Problem Solving

705-805

Sequences

The sequence a1, a2, a3, ..., an,…...is such that $a_n = \frac{1}{n^2} - \frac{1}{(n-1)^2}$ for all integers $n \geq 2$ and $a_1 = 1$ . Which of the following could be the sum of sequence if the number of terms in the sequence is a prime number?

Answer Choices

Correct answer marked below

A

\frac{1}{13}

B

\frac{1}{31}

C

\frac{1}{225}

D

\frac{1}{289}

\frac{1}{324}

Correct

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Official Explanation

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Given:

a_n = \frac{1}{n^2} - \frac{1}{(n-1)^2}

for

n \geq 2

, and

a_1 = 1

.

Let the sequence have

k

terms, where

k

is a prime number.

The sum

S_k = a_1 + a_2 + a_3 + \dots + a_k

.

Substitute the formula for

a_n

:

S_k = 1 + \left( \frac{1}{2^2} - \frac{1}{1^2} \right) + \left( \frac{1}{3^2} - \frac{1}{2^2} \right) + \dots + \left( \frac{1}{k^2} - \frac{1}{(k-1)^2} \right)

.

Observe that this is a telescoping series. Most terms cancel:

S_k = 1 - 1 + \frac{1}{2^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{3^2} + \dots + \frac{1}{(k-1)^2} - \frac{1}{(k-1)^2} + \frac{1}{k^2}

.

All intermediate terms cancel, leaving:

S_k = \frac{1}{k^2}

.

Thus, the sum of the first

k

terms is

\frac{1}{k^2}

, where

k

is the number of terms (and is given to be a prime number).

We check each option to see if it equals

\frac{1}{p^2}

for some prime

p

:

A.

\frac{1}{13}

→

13

is prime, but

\frac{1}{13} \neq \frac{1}{p^2}

since

13

is not a perfect square.

B.

\frac{1}{31}

→

31

is prime, but

31

is not a perfect square.

C.

\frac{1}{225}

→

225 = 15^2

, but

15

is not prime.

D.

\frac{1}{289}

→

289 = 17^2

, but

17

is prime. However, note: the sum is

\frac{1}{k^2}

where

k

is the number of terms. Here,

k = 17

(prime) would give sum

\frac{1}{289}

. This is a candidate.

E.

\frac{1}{324}

→

324 = 18^2

, but

18

is not prime.

Wait: Check option D again:

\frac{1}{289} = \frac{1}{17^2}

. If the sequence has 17 terms (17 is prime), the sum is

\frac{1}{17^2} = \frac{1}{289}

The sequence a1, a2, a3, ..., an,…...is such that $a_n = \frac{1}{n^2} - \frac{1}{(n-1)^2}$ for all integers $n \geq 2$ and $a_1 = 1$ . Which of the following could be the sum of sequence if the number of terms in the sequence is a prime number?

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