The sum of all positive integers n for which 13-23-2n3-12-22-n2 is also an integer is-

1^3+2^3....+(2n)^3/1^2+2^2+...n^2 = (2n(2n+1)^2/2 nbsp; )^2 6/n(n+1)(2n+1) =6n(2n+1)/n+1 =12n^2+6n/n+1=12(n^2+1)+6(n+1)+6/n+1 = 12n 6 + 6/n+1 If the g ...

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Standard X

Mathematics

Solving Inequalities

The sum of al...

Question

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

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Solution

The correct option is A
8

$\frac{1^{3} + 2^{3} . . . . + (2 n)^{3}}{1^{2} + 2^{2} + . . . n^{2}} = {(\frac{2 n (2 n + 1)^{2}}{2})}^{2} \frac{6}{n (n + 1) (2 n + 1)}$

$= \frac{6 n (2 n + 1)}{n + 1}$

$= \frac{12 n^{2} + 6 n}{n + 1} = \frac{12 (n^{2} + 1) + 6 (n + 1) + 6}{n + 1}$

$= 12 n - 6 + \frac{6}{n + 1}$

If the given terms are an integers, then $\frac{6}{n + 1}$ must be an integer

$\Rightarrow n = 1, 2, 5$

$S u m = 8$

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The sum of all positive integers n for which 13+23....+(2n)312+22+...n2 is also an integer is.

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