The ratio of squares of first n natural numbers to square of sum of first n natural numbers is17:325. The value of n is:

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Solution

The correct option is B
25

Sum of squares of first n natural numbers= $\frac{n (n + 1) (2 n + 1)}{6}$

Sum of first n natural numbers = $\frac{n (n + 1)}{2}$

So solving the given equation we will get n=25

Suggest Corrections

Similar questions

The remainder obtained when the sum of squares first n odd natural numbers is divided the sum of squares of first n even natural numbers is

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

If sum of the squares of first n natural numbers exceeds their sum by 330, then find the value of n.

The difference between the sum of the cubes of first n natural numbers and squares of first n natural numbers is

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\frac{n^{2} - 1}{6}

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n^{2}

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\frac{n}{3} (4 n^{2} - 1)

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