Question

Find the sum of $3^2+4^2+5^2+....$ up to n terms.

Answer 1

$3^2+4^2+5^2+.....$ up to n terms = ?

We know that sum of squares of first n natural

numbers is $\frac{n(n+1)(2n+1)}{6}$

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

$1^2+2^2+$

$\Rightarrow 3^2+4^2+....+(n+2{)}^2=\frac{(n+2)(n+3)(2n+5)}{6}$

$n^{th}$ term in given series

$\Rightarrow 3^2+4^2+.....$ upto n terms

$=\frac{n(n+2)(n+3)(2n+5)}{6}-5$