Question

Find the sum of n terms of the series. Whose $n^{th}$ terms is given
(i) $n(n+1)(n+4)$
(ii) $n^2+2^n$

Answer 1

$\left(i\right)$

$a_n=n(n+1)(n+4)$

$=>(n^2+n)(n+4)$

$=>n^3+5n^2+4n$

then sum of $n$ terms

$s_n=\sum _{n\to 1}(n^3+5n^2+4n)$

$=>\sum x\to 1(n^3+\sum _{x\to z}(5n^2)+\sum x\to 1(n\left)\right)$

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

$=>n(n+1)\times \left[\frac{3n(n+1+10(2n+1)+2n)}{12}\right]$

$\left(ii\right)$

$=>s_n=\sum _{x\to 1}\left(a_n\right)$

$=>\sum _{x\to 1}(n^2+2^n)$

$\sum _{x\to 1}\left(n^2\right)+\sum _{x\to 1}\left(2^n\right)$

Now,

$=>\sum _{x\to 1}\left(2^n\right)-----------------\left(i\right)$

$=>2^1+2^2+2^3+.......{.2}^n$

This is in GP, with first term $a=2$ and common ratio $r=\frac{4}{2}=2$

$=>s_n=\frac{a(r^n-1)}{r-1}$

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

$\sum _{x\to 1}\left(2^n\right)=>2(2^n-1)$

Now from $\left(i\right)$ we get

$=>s_n=\sum _{x\to 1}\left(n^2\right)-\sum _{x\to 1}\left(2^n\right)$

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