Question:
If $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}=210$, find the value of $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{2}$.
Solution:
It is given that, $\sum_{k=1}^{n} k=210$
Note:
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,
$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}=210$
$n^{2}+n=420$
$n^{2}+n-420=0$
$(n-20)(n+21)=0$
$n=20$ 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, $\sum_{k=1}^{n} k^{2}=\frac{20(21)(41)}{6}=2870$