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Answer 1 · 2017-10-13T20:00:31

The answer is $n=29$

To solve this, we can use a very known result, which is the formula to the sum of the $n$ first squares. It is as follows:

$\sum_[k=1]^n k^2 = (n(n+1)(2n+1))/6$

Substituting the formula into the equation, we get

$1/(n(2n+1)) * (n(n+1)(2n+1))/6 = 5$

Simplifying:

$(n+1)/6 = 5$

All we need to do now is solve for $n$

$n+1 = 30 => n=29$ //

Proof to the formula of the sum of the $n$ first squares can be found here: http://mathforum.org/library/drmath/view/56920.html

Hope it helps.