Find the sum $11^{2}+12^{2}+13^{2}+\ldots 20^{2}$
It is required to find the sum $11^{2}+12^{2}+13^{2}+\ldots 20^{2}$
$11^{2}+12^{2}+13^{2}+\ldots 20^{2}=$ Sum of squares of natural numbers starting from 1 to $20-$ Sum of squares of natural numbers starting from 1 to 10 .
Note:
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 of squares of natural numbers starting from 1 to 20
$=\frac{20(21)(41)}{6}=2870$
Sum of squares of natural numbers starting from 1 to 10
$=\frac{10(11)(21)}{6}=385$
$11^{2}+12^{2}+13^{2}+\ldots 20^{2}=$ Sum of squares of natural numbers starting from 1 to $20-$ Sum of squares of natural numbers starting from 1 to 10 .
$11^{2}+12^{2}+13^{2}+\ldots 20^{2}=2870-385=2485$