The sequence a1, a2, a3, ..., an,…...is such that an= 1/n^2−1/(n−1)^2

Question

The sequence a1, a2, a3, ..., an,…...is such that an= 1/n^2−1/(n−1)^2 for all integers n≥2 and a1 = 1. Which of the following could be the sum of sequence if the number of terms in the sequence is a prime number?

A. 1/13
B. 1/31
C. 1/225
D. 1/289
E. 1/324

A. 1/13
B. 1/31
C. 1/225
D. 1/289
E. 1/324

chetan2u · Answer

a_n= \frac{1}{n^2}−\frac{1}{(n−1)^2} ...
 a_1=1 
 a_2= \frac{1}{2^2}−\frac{1}{(2−1)^2} ...
As 2 is prime, the sum is  a_1+a_2=1+ \frac{1}{2^2}−\frac{1}{(2−1)^2}=\frac{1}{2^2} ...
 a_3= \frac{1}{3^2}−\frac{1}{(3−1)^2} ...
As 3 is prime, the sum is  a_1+a_2+a_3=\frac{1}{2^2}+\frac{1}{3^2}−\frac{1}{(3−1)^2}=\frac{1}{3^2} ...

So, the sum is  \frac{1}{n^2} , where n is the number of terms..

Since we are taking number of terms as prime number, let us check each option for the denominator as square of a prime number..
A. 1/13....NO
B. 1/31....NO
C. 1/225= \frac{1}{15^2} ..15 is not prime....NO
D. 1/289= \frac{1}{17^2} ..17 is prime....YES
E. 1/324= \frac{1}{9*36}=\frac{1}{18^2} ..18 is not prime....NO

D

Fdambro294 · Answer

In most sequence questions, it seems as if the key is to understand the pattern that is occurring.

You can usually do that by looking at the first 4 or 5 terms and looking for a pattern in:

-the terms themselves

-the difference between consecutive terms

-or sometimes in the addition of consecutive terms

In this problem, keeping the terms together and not performing the calculations for each term helps to see what is actually going on.

A1 = 1

A2 = (1/4) - 1

If we add up A1 and A2 right now, the (1 / (n - 1)^2) term in A2 will cancel out with A1

A3 = (1/9) - (1/4)

Again, if we sum up all the terms here, we can see that the (1 / (n - 1)^2) term in A3 will cancel out with the (1 / n^2) term in the prior A2

In fact this will keep happening as you add up any series of terms. All that will remain is the (1 / n^2) term for that given An

A4 = (1/16) - (1/9)

Sum up through A4:

1 + (1/4) - 1 + (1/9) - (1/4) + (1/16) - (1/9) =

1/16

Which is the term (1 / n^2) for the given term we added up through ——-> A4

Thus, if we are summing up a series of terms through a Prime Number - An

We are looking for an answer that will give us:

(1 / n^2) ——> where the DEN is the Perfect Square of a Prime Number

Only D gives us that option.

1 / (17)^2 = 1/289

D

Posted from my mobile device

mimishyu · Answer

a1+a2+a3+a4..... an
=(1/1^2 - 1/0^2) + (1/2^2 - 1/1^2) + (1/3^2 - 1/2^2) + (1/4^2 - 1/3^2)..... (1/n^2 - 1/(n-1)^2)
cancel those two terms that's same in their base but are different
in their coefficient, in which, one is positive +1 while the other is
negative -1
thus the sum to the sequence of n will be 1/n^2
to square root the numbers among choices (A)-(E), 
we get 289=17^2, to which 17 is a prime number
the correct answer is (D)

CEdward · Answer

an= 1/n^2−1/(n−1)^2 is terribly confusing.
This should be rewritten ...

an = (1/n^2) - (1/(n-1)^2)

bumpbot · Answer

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The sequence a1, a2, a3, ..., an,…...is such that an= 1/n^2−1/(n−1)^2

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The sequence a1, a2, a3, ..., an,…...is such that an= 1/n^2−1/(n−1)^2

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