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Calculus Level 3

1 2 x 0 + 2 2 x 1 + 3 2 x 2 + 4 2 x 3 + ⋯ \large 1^2x^0 + 2^2x^1+3^2x^2+4^2x^3+ \cdots

Find the closed form of the series above for ∣ x ∣ < 1 |x| < 1 .

1 − x ( 1 + x ) 2 \dfrac{1-x}{(1+x)^2} 1 − x ( 1 + x ) 3 \dfrac{1-x}{(1+x)^3} 1 + x ( 1 − x ) 3 \dfrac{1+x}{(1-x)^3} 2 + x ( 1 − x ) 3 \dfrac{2+x}{(1-x)^3}

Sabhrant Sachan by Sabhrant Sachan , 21, India

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1 solution

Rudraksh Shukla
May 1, 2016

I went for an algebraic approach, S = 1 2 x 0 + 2 2 x 1 + 3 2 x 2 ⋯ S=1^2x^0+2^2x^1+3^2x^2 \cdots Multiply S by x x and subtract from original series, we shall have S ( 1 − x ) = 1 + 3 x + 5 x 2 + 7 x 3 ⋯ S(1-x)=1+3x+5x^2+7x^3\cdots which is an arithmetico geometric progression. Again I multiply x x to S ( 1 − x ) S(1-x) and then subtract it from S ( 1 − x ) S(1-x) , then, S ( 1 − x ) 2 = 1 + 2 x + 2 x 2 + 2 x 3 ⋯ S(1-x)^2=1+2x+2x^2+2x^3\cdots ⇒ S ( 1 − x ) 2 = 1 + 2 ( x + x 2 + x 3 ⋯ ) \Rightarrow S(1-x)^2=1+2(x+x^2+x^3\cdots)

⇒ S ( 1 − x ) 2 = 1 + 2 x 1 − x \Rightarrow S(1-x)^2=1+\dfrac{2x}{1-x} ⇒ S = 1 + x ( 1 − x ) 3 \Rightarrow S=\dfrac{1+x}{(1-x)^3}

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