The sequence a1, a2, a3, ..., an, ... is such that an=1/n^21/(n1)^2

Question

The sequence  a_1, \ a_2, \ a_3, \ ..., \ a_n, \ ...  is such that  a_n=\frac{1}{n^2}−\frac{1}{(n−1)^2}  for all integers  n≥2  and  a_1 = 1 . Is the sum of this sequence greater than 0.005?

(1) The number of terms in the sequence is greater than 12

(2) The number of terms in the sequence is lesser than 14

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truplayer257 · Answer

a1 = 1
a2 = 1/(2^2) - 1/(2 - 1)^2 = 1/4 - 1
a3 = 1/(3^2) - 1/4 = 1/9 - 1/4

a1 + a2 + a3 + ....
1 + 1/4 - 1 + 1/9 - 1/4

Do you see the pattern here? Basically, the 2nd term of the expression 1/(n^2) - 1/(n - 1)^2 is cancelling out. So following this logic:

Statement 1:
Since n > 12, the sum has to be less than 1/(12^2) or 1/144. Note that 0.005 is 1/200, so this can or cannot be true if n > 12. Insufficient.

Statement 2:
Since n < 14, and 1/(14^2) = 1/196 > 1/200, the sum of the sequence must be greater than 1/200. 
Sufficient.

Answer is B

Trent981 · Answer

Hi  Bunuel ,  chetan2u

Can you kindly help me understand why the answer is C and not B ?

thanks

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} ...
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} ...
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.

Now,  0.005=\frac{5}{1000}=\frac{1}{200}  and  \frac{1}{14^2}>\frac{1}{200}>\frac{1}{15^2} 
So, we are checking whether the number of terms is less than 15 OR greater than or equal to 15.

(1) The number of terms in the sequence is greater than 12
So, n could be 13 or 14, then the answer would be YES. But 15 and above would give answer as NO. 
Insufficient

(2) The number of terms in the sequence is lesser than 14
So, n is always less than 15, and the answer would be YES always. 
Sufficient

B

Trent981 , I believe OA has been marked C erroneously.

bumpbot · Answer

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The sequence a1, a2, a3, ..., an, ... is such that an=1/n^21/(n1)^2

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GMAT Data Sufficiency (DS) Questions

The sequence a1, a2, a3, ..., an, ... is such that an=1/n^21/(n1)^2

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