Question

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

Answer 1

Given,

$T_n=3^n-2^n$

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

We know sum of n terms of a G.P. is

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

where, a = first term of G P.

r = common ratio of the G.P.

Now, $\Sigma 3^n=\frac{3(3^n-1)}{3-1}$ [a = 3 , r = 3]

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

Similarly, $\Sigma 2^n=\frac{2(2^n-1)}{2-1}$ [a =2, r =2]

$=2^{n+1}-2$ -------------- (3)

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

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

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