An algebra problem by A Former Brilliant Member

Algebra Level 3

S = ∑ x = 2 ∞ 3 x 2 + 1 ( x 2 − 1 ) 3 S = \displaystyle \sum _{ x=2 }^{ \infty }{ \dfrac { { 3 }{ x }^{ 2 }+{ 1 } }{ { \left( { x }^{ 2 }-{ 1 } \right) }^{ 3 } } }

For S S as given above, what is the value of 9 S \dfrac { 9 }{ S } ?


The answer is 16.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Chew-Seong Cheong
Mar 24, 2018

S = ∑ x = 2 ∞ 3 x 2 + 1 ( x 2 − 1 ) 3 = ∑ x = 2 ∞ 3 x 2 + 1 ( x − 1 ) 3 ( x + 1 ) 3 = 1 2 ∑ x = 2 ∞ ( 1 ( x − 1 ) 3 − 1 ( x + 1 ) 3 ) = 1 2 ( ∑ x = 1 ∞ 1 x 3 − ∑ x = 3 ∞ 1 x 3 ) = 1 2 ( 1 1 3 + 1 2 3 ) = 9 16 \begin{aligned} S & = \sum_{x=2}^\infty \frac {3x^2+1}{(x^2-1)^3} \\ & = \sum_{x=2}^\infty \frac {3x^2+1}{(x-1)^3(x+1)^3} \\ & = \frac 12 \sum_{x=2}^\infty \left( \frac 1{(x-1)^3} - \frac 1{(x+1)^3}\right) \\ & = \frac 12 \left(\sum_{x=1}^\infty \frac 1{x^3} - \sum_{x=3}^\infty \frac 1{x^3} \right) \\ & = \frac 12 \left(\frac 1{1^3} + \frac 1{2^3} \right) \\ & = \frac 9{16} \end{aligned}

Therefore, 9 S = 9 9 16 = 16 \dfrac 9S = \dfrac 9{\frac 9{16}} = \boxed{16} .

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