# The infinite sum, $1+\frac{4}{7}+\frac{9}{7^{2}}+\frac{16}{7^{3}}+\frac{25}{7^{4}}+$ ----- equals

This is a simple example of a series called Arithmetic-Geometric Progression.

Let the sum that you need to find be S.

$S=1+\frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+..$  (eqn 1)

Multiply S by the common ratio (in this case  1/7  )

So, we get,

$\frac{S}{7}=\frac{1}{7}+\frac{4}{7^2}+\frac{9}{7^3}+\frac{16}{7^4}+..$ ..  (eqn 2)

Subtract equation 2 from 1,

$S(1-\frac{1}{7})= 1+ \frac{(4-1)}{7}+\frac{(9-4)}{7^2} + \frac{(16-9)}{7^3}+....$

$S(\frac{6}{7})= 1+ \frac{(3)}{7}+\frac{(5)}{7^2} + \frac{(7)}{7^3}+....$

We now see that the numerators on the right hand side form an arithmetic progression. For solving this type of a problem, multiply both sides by the common ratio again ( 1/7 ) and subtract.

$S\left(\frac{6}{7}\right)\left(1-\frac{1}{7}\right)= 1+ \frac{(3-1)}{7}+\frac{(5-3)}{7^2} + \frac{(7-5)}{7^3}+....$

$S(36/49)=1+2[1/7+1/7^2+1/7^3+...]$

The terms in the bracket form a simple infinite geometric progression.

(Refer Geometric progression if you are unaware of the formula)

S(36/49)=1+2[(1/7)/(1−1/7)]

S(36/49)=4/3

S=49/27

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