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If $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}=210$, find the value of $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{2}$.



It is given that, $\sum_{k=1}^{n} k=210$


I. Sum of first n natural numbers, 1 + 2 +3+…n,

$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$

II. Sum of squares of first $n$ natural numbers, $1^{2}+2^{2}+3^{2}+\ldots n^{2}$,

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

From the above identities,

$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}=210$




$n=20$ or $-21$

Since, n is the number of integers, n = 20

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

So, $\sum_{k=1}^{n} k^{2}=\frac{20(21)(41)}{6}=2870$

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