Question

Find the sum of $n$ terms of the series whose $n^{th}$ term is
$3(4^n+2n^2)-{4n}^3$.

Answer 1

The general term is given by,

$T_n=3(4^n+2n^2)-4n^3$

$=3\times 4^n+6n^2-4n^3$

Now, $\Sigma T_n=3\Sigma 4^n+6\Sigma n^2-4\Sigma n^3$ ------------ (1)

Now, sum of n terms of G.P.

$S_n=\frac{a(r^n-1)}{r-1}$ ------------ (2)

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

$\Sigma n^3=\frac{n^2(n+1{)}^2}{4}$ ------------- (4)

Using (2),(3) and (4) in equation (1),

$\Sigma T_n=3\times \frac{4(4^n-1)}{4-1}+6\times \frac{n(n+1)(2n+1)}{6}-4\times \frac{n^2(n+1{)}^2}{4}$

$=4(4^n-1)+n(n+1)(2n+1)+n^2(n+1{)}^2$

$=4^{n+1}-4+2n^3+3n^2+n+n^4+2n^3+n^2$

$\Sigma T_n=n^4+4n^3+4n^2+n-4+4^{n+1}$