Question

Find the sum to infinite terms of the series
1 + 5x + 9x2 + 13x? +.....o where
| < 1​

Answer 1

Given :

A.G series

1 + 5x + 9x² + 13x³ + ...

To Find :

Sum of infinite terms of the series

Solution :

Sum of infinite terms of A.G ( Arithmetic - Geometric ) series is given by ,

$\bigstar\ \; \sf \pink{S_{\infty}=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}}$

Sum of infinite terms of G.P ( Geometric ) series is given by ,

$\sf \bigstar\ \; \green{S_{\infty}=\dfrac{a}{1-r}}$

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Given ,

a = 1 , r = x , d = 5 - 1 = 4

Sub. values ,

$\sf \to S_{\infty}=\dfrac{1}{1-x}+\dfrac{4x}{(1-x)^2}\\\\\to \sf S_{\infty}=\dfrac{1-x+4x}{(1-x)^2}\\\\\to \sf S_{\infty}=\dfrac{1+3x}{(1-x)^2}\ \; \bigstar$

Alternate Method :

Let ,

S = 1 + 5x + 9x² + 13x³ + ... ___ (1)

Multiply both sides with ' x ' ,

⇒ Sx = x + 5x² + 9x³ + 13x⁴ + ... ___ (2)

Solve (2) - (1) ,

⇒ S - Sx = ( 1 + 5x + 9x² + 13x³ + ... ) -

( x + 5x² + 9x³ + 13x⁴ + ... )

⇒ S ( 1 - x ) = 1 + ( 4x + 4x² + 4x³ + ... )

⇒ S ( 1 - x ) = 1 + $\sf \bigg(\dfrac{4x}{1-x}\bigg)$

⇒ S = $\sf \dfrac{1+3x}{(1-x)^2}$