# infinite sum

• Jun 20th 2010, 01:08 AM
jashansinghal
infinite sum
the infinite sum 1+ 4/7 + 9/7^2 +16/7^3 + 25/7^4 + ...... equals?
• Jun 20th 2010, 02:31 AM
simplependulum
I prefer to use techniques of differentiation :

$\displaystyle \frac{1^1}{7^0} + \frac{2^2}{7} + \frac{3^2}{7^2} + \frac{4^2}{7^3} + ...$

Consider

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + ....$

Take differential ,

$\displaystyle \frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + .. ~~~~~ -[1]$

Take it agian ,

$\displaystyle \frac{2}{(1-x)^3} = 2 + 3(2) x + 4(3) x^2 + ... ~~~~~ -[2]$

$\displaystyle [1] + x [2]$

$\displaystyle \frac{1}{(1-x)^2} + \frac{2x}{(1-x)^3} = 1 + (2+2)x + (3+3(2))x^2 + (4 + 4(3))x^3 + ...$

$\displaystyle = 1+2^2x + 3^2x^2 + 4^2 x^3 + ....$

Sub . $\displaystyle x = \frac{1}{7}$

$\displaystyle \frac{1^1}{7^0} + \frac{2^2}{7} + \frac{3^2}{7^2} + \frac{4^2}{7^3} + ... = \frac{1}{(1-1/7)^2} + \frac{2/7}{(1-1/7)^3}$

$\displaystyle = \frac{49}{36} + \frac{49}{108}$

$\displaystyle = \frac{49}{27}$