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Algebra Level 3

1 ( 1 + x ) ( 1 + 2 x ) + 1 ( 1 + 2 x ) ( 1 + 3 x ) + 1 ( 1 + 3 x ) ( 1 + 4 x ) + ⋯ \large \dfrac{1}{(1+x)(1+2x)}+\dfrac{1}{(1+2x)(1+3x)}+\dfrac{1}{(1+3x)(1+4x)} + \cdots

Find the closed form of the sum of the series above up to n n terms.

n ( 1 + x ) [ 1 + ( n 2 + 1 ) x ] \frac{n}{(1+x)[1+(n^2+1)x]} n ( 1 + x ) [ 1 + ( n + 1 ) x ] \frac{n}{(1+x)[1+(n+1)x]} n + 1 ( 1 + x ) [ 1 + ( n + 1 ) x ] \frac{n+1}{(1+x)[1+(n+1)x]} n 2 ( 1 + x ) [ 1 + ( n + 1 ) x ] \frac{n^2}{(1+x)[1+(n+1)x]}

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Rudraksh Shukla by Rudraksh Shukla , 23, India

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1 solution

Aditya Dhawan
Apr 7, 2016

Relevant wiki: Telescoping Series - Sum

T h e g i v e n s e r i e s i s e q u i v a l e n t t o : ∑ k = 1 n 1 [ 1 + k x ] [ 1 + ( k + 1 ) x ] = 1 x ∑ 1 + ( 1 + k ) x − [ 1 + k x ] [ 1 + k x ] [ 1 + ( k + 1 ) x ] = 1 x ∑ 1 1 + k x − 1 1 + ( k + 1 ) x ( T E L E S C O P I N G S E R I E S ) = 1 x ( 1 x + 1 − 1 1 + ( n + 1 ) x ) = n ( x + 1 ) ( 1 + ( n + 1 ) x ) The\quad given\quad series\quad is\quad equivalent\quad to:\\ \sum _{ k=1 }^{ n }{ \frac { 1 }{ [1+kx][1+(k+1)x] } } \\ =\frac { 1 }{ x } \sum { \frac { 1+(1+k)x-[1+kx] }{ [1+kx][1+(k+1)x] } } \\ =\frac { 1 }{ x } \sum { \frac { 1 }{ 1+kx } } -\frac { 1 }{ 1+(k+1)x } \quad (TELESCOPING\quad SERIES)\\ =\frac { 1 }{ x } \left( \frac { 1 }{ x+1 } -\quad \frac { 1 }{ 1+(n+1)x } \right) =\frac { n }{ (x+1)(1+(n+1)x) } \\

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