Question

Plz help me !!

Sum of n terms of series

4 +24+96+320+..............​

Answer 1

Answer:

$\bold\red{S_{n} = ( {n}^{2} - n + 2) {2}^{n + 1} - 4}$

Step-by-step explanation:

Given,

A series,

$4 + 24 + 96 + 320 + ..........$

Now,

we can also simplify the given series as,

$(2 \times 2 )+ (6 \times 4 )+ (12 \times 8 )+ (20 \times 16) + ...........$

After observation in the series,

2, 6, 12, 20, 30, .............

we get,

The nth term is given by,

${n}^{2} + n$

Hence,

for the given series,

we have ,

the nth term is given by ,

$\bold{u_{n} = ( {n}^{2} + n) {2}^{n}}$

Now,

Let's assume that,

$( {n}^{2} + n) {2}^{n} = (A {n}^{2} + Bn + C) {2}^{n} - (A {(n - 1)}^{2} + B(n - 1) + C) {2}^{n - 1}$

Now,

Dividing out by ${2}^{n-1}$ and equatting coefficients of like powers of n,

we get,

$A = 2 \\ \\ 2A+ B= 2 \\ \\ C - A + B = 0$

From this,

we get,

$= > A= 2 \\ \\ = > B = - 2 \\ \\ = > C = 4$

Therefore,

further we get,

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

Hence,

Sum of the series,

$= > S_{n} = (2 {n}^{2} - 2n + 4) {2}^{n} - 4 \\ \\ = > \bold{S_{n} = ( {n}^{2} - n + 2) {2}^{n + 1} - 4}$