Question

$1^2+{\left(\frac{3}{2}\right)}^2+2^2+{\left(\frac{5}{2}\right)}^2+...$ to $40$ terms.

Answer 1

Given series is $1^2+{\left(\frac{3}{2}\right)}^2+2^2+{\left(\frac{5}{2}\right)}^2+......+$ to $40$ terms

$1^2=1^2,{\left(\frac{3}{2}\right)}^2={(1+\frac{1}{2}(2-1\left)\right)}^2$

$(2{)}^2={(1+\frac{1}{2}(3-1\left)\right)}^2$......

So, the general $n^{th}$ term is $t_n={(1+\frac{1}{2}(n-1\left)\right)}^2$

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

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

$=\frac{n^2+2n+1}{4}$

$\therefore$ Sum of n terms, $s_n=\frac{1}{4}.\frac{n(n+1)(2n+1)}{6}+\frac{2n(n+1)}{2.4}+\frac{n}{4}$

So, $S_{40}=\frac{1}{4}.\frac{\mathrm{40.41.81}}{6}+\frac{40.41}{4}+\frac{40}{4}$

$=5535+410+10$

$=5955$