Question

Find the sum of n terms of the series in Excercise 8 to 10 whose $n^{th}$ terms is given
1.) $\left(n+1\right)\left(n+4\right)$
2.) $n^2+2^n$

Answer 1

$\left(i\right)$Given:$T_n=(n+1)(n+4)$

$=n^2+4n+n+4$

$=n^2+5n+4$

$S_n=\sum _{n=1}^n(n^2+5n+4)$

$=\sum _{n=1}^n{n}^2+5\sum _{n=1}^{n}n+4\sum _{n=1}^{n}1$

$=\frac{n(n+1)(2n+1)}{6}+5\times \frac{n(n+1)}{2}+4n$

$=\frac{n(n+1)(2n+1)}{6}+\frac{5n(n+1)}{2}+4n$

$=n[\frac{(n+1)(2n+1)}{6}+\frac{5n+5}{2}+4]$

$=\frac{n}{6}[(n+1)(2n+1)+15n+15+24]$

$=\frac{n}{6}[2n^2+n+2n+1+15n+15+24]$

$=\frac{n}{6}[2n^2+18n+40]$

$=\frac{2n}{6}(n^2+9n+20)$

$=\frac{n}{3}(n^2+4n+5n+20)$

$=\frac{n}{3}(n(n+4)+5(n+4))$

$=\frac{n(n+4)(n+5)}{3}$

$\left(ii\right)$Given:$t_n=n^2+2^n$

$S_n={\sum }_{n=1}^n{t}_n$

$={\sum }_{n=1}^n{n}^2+2^n$

$={\sum }_{n=1}^n{n}^2+{\sum }_{n=1}^n{2}^n$

$=\frac{n(n+1)(2n+1)}{6}+[2+4+8+...+2^n]$ ........$\left(1\right)$

Consider $2+4+8+...+2^n$ where $a=2,commonratio=r=\frac{4}{2}=2$

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

Substituting in $\left(1\right)$ we get

${\sum }_{n=1}^n{n}^2+{\sum }_{n=1}^n{2}^n$

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