Question

a÷1+i +a÷(1+i)^2+a÷(1+i)^3+...+a÷(1+i)^n find the sum in gp

Answer 1

Answer:

The required sum is $S_n=(\frac{a}{1+i})(\frac{((1+i)^n-1)(1+i)^{1-n}}{i})$

Step-by-step explanation:

Given : $\frac{a}{1+i}+\frac{a}{(1+i)^2}+\frac{a}{(1+i)^3}+....+\frac{a}{(1+i)^n}$

To find : The sum of G.P?

Solution :

The Geometric series is in the form $a,ar,ar^2,ar^3,...,ar^{n-1}$

Where, a is the first term and r is the common ratio.

The sum of finite G.P is given by,

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

In the given series, $\frac{a}{1+i}+\frac{a}{(1+i)^2}+\frac{a}{(1+i)^3}+....+\frac{a}{(1+i)^n}$

Here, $a=\frac{a}{1+i}$ and  $r=\frac{1}{1+i}$

Substitute in the formula,

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

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

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

Therefore, The required sum is $S_n=(\frac{a}{1+i})(\frac{((1+i)^n-1)(1+i)^{1-n}}{i})$