Question

Find the sum to infinite terms of the series 1+5x+9x^2+13x^3+..... Where |x| <1

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Series is}$

$\mathsf{1+5x+9x^2+13x^3+.......}$

$\underline{\textsf{To find:}}$

$\textsf{Sum to infinite of the given series}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{1+5x+9x^2+13x^3+.......}$

$\textsf{clearly, it is an arithmetico-geometric series}$

$\textsf{with a=1, r=x and d=4}$

$\textsf{Then,}$

$\textsf{Sum to infinite terms of the series}$

$\mathsf{S_{\infty}=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}}$

$\mathsf{S_{\infty}=\dfrac{1}{1-x}+\dfrac{4x}{(1-x)^2}}$

$\mathsf{S_{\infty}=\dfrac{1-x+4x}{(1-x)^2}}$

$\implies\boxed{\mathsf{S_{\infty}=\dfrac{3x+1}{(1-x)^2}}}$

$\underline{\textsf{Answer:}}$

$\textsf{The sum t infinite terms of the given series is}\;\mathsf{\dfrac{3x+1}{(1-x)^2}}$

Answer 2

SOLUTION

TO DETERMINE

The sum to infinite terms of the series

$\sf{ 1+ 5x + 9 {x}^{2} + 13 {x}^{3} + ........ \: \: }$

$\sf{ \: where \: \: |x| < 1\: }$

EVALUATION

Let

$\sf{S = 1+ 5x + 9 {x}^{2} + 13 {x}^{3} + ........ } - - - - (1)$

Multiplying both sides of Equation (1) by x we get

$\sf{Sx = x+ 5 {x}^{2} + 9 {x}^{3} + 13 {x}^{4} + ........ } - - - - (2)$

Equation (1) - Equation (2) gives

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

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

Now

$\sf{ 4 + 4x+ 4{x}^{2} + 4 {x}^{3} + ........ }$

is a Geometric infinite series with first term = 4 and Common ratio = x with |x|< 1

So

$\displaystyle \sf{S(1 - x ) + 3= \frac{4}{1 - x} }$

$\implies\displaystyle \sf{S(1 - x )= \frac{4}{1 - x} - 3}$

$\implies\displaystyle \sf{S(1 - x )= \frac{4 - 3 + 3x}{1 - x} }$

$\implies\displaystyle \sf{S(1 - x )= \frac{ 3x + 1}{1 - x} }$

$\implies\displaystyle \sf{S= \frac{ 3x + 1}{ {(1 - x)}^{2} } }$

FINAL ANSWER

$\displaystyle \sf{1+ 5x + 9 {x}^{2} + 13 {x}^{3} + .....= \frac{ 3x + 1}{ {(1 - x)}^{2} } }$

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