The sum of all positive integers n for which \frac{1^ 3 + 2^3 +....+ (2n)^3}{1^2 + 2^2+.... + n^2} is also an integer is

KVPY SA Exam 2014 Question Paper

Question : 3 of 80

Marks: +1, -0

The sum of all positive integers n for which

{1^ 3 + 2^3 +....+ (2n)^3}/{1^2 + 2^2+.... + n^2}

is also an integer is

Solution:

{({2n(2n +1)}/2)^2}/{{n(n +1)(2n +1)}/6}= {6n ^2(2n+ 1) ^2 }/{n(n +1)(2n +1)}

{6n (2n+ 1) }/(n+1)={6n(n+ 1+ n)}/{n+1}

=6n+{6n^2}/{n+1}

= 6n + 6(n - 1) + 6/{n+1}=

Integer
The values of 'n' which are satisfying are n = 1, 2, 5 only for being integer
∴ sum = 1 + 2 + 5 = 8

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