Given series is $5^2+6^2+7^2+........20^2$

We have to find the $n^{th}$ term of this series.

Each term in this series is square of a number.

$i.e.,$ The $n^{th}$ term of the series is square of the $n^{th}$ term

of the sequence $5,6,7...........$

This sequence is an A.P. whose $n^{th}$ term is $5+(n-1).1$

Since, the $n^{th}$ term of an $A.P.$ is $a+(n-1)d$

$\Rightarrow\:4+n=20$ or $n=16$

$i.e., $ The number of terms in the given series is 1$n=16$

$\therefore$ The $n^{th}$ term of the given series is $t_n=(4+n)^2$

$\Rightarrow\:t_n=16+n^2+8n$

Step 2

We have to find the sum of 16 terms of the given series =$S_{16}$

The sum of $n$ terms of the any series is $S_n=\sum \ t_n$

$=\sum (16+8n+n^2)$

$=\sum 16+8\sum n+\sum n^2$

We know that $\sum n=\large\frac{n(n+1)}{2}$ and

$\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$

Also we know that $\sum k=kn$

$\Rightarrow \sum 16=16n$

$8\sum n=8\times \large\frac{n(n+1)}{2}$$=4n(n+1)$

$\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$

But since we have to find the sum of 16 terms only.

$\therefore n=16$

Substituting the value of $n$ we get

$\sum 16=16\times 16=256$

$8\sum n=4\times 16\times (16+1)=64\times 17=1088$

and

$\sum n^2=\large\frac{16\times (16+1)(2\times 16+1)}{6}$

$=\large\frac{16\times 17\times 33}{6}$$=1496$

$\Rightarrow\:S_{16}=256+1088+1496=2840$