The infinite summation

Algebra Level 4

∑ k = 1 ∞ 6 k ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) \large \sum_{k=1}^{\infty} \dfrac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}

Evaluate the sum above.


The answer is 2.

Vilakshan Gupta by Vilakshan Gupta , 19, India

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3 solutions

Mark Hennings
Apr 4, 2018

Since 6 k ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) = 2 k 3 k − 2 k − 2 k + 1 3 k + 1 − 2 k + 1 \frac{6^k}{(3^k - 2^k)(3^{k+1}-2^{k+1})} \; = \; \frac{2^k}{3^k - 2^k} - \frac{2^{k+1}}{3^{k+1} - 2^{k+1}} the series telescopes, and the sum is 2 3 − 2 = 2 \frac{2}{3-2} \; = \; \boxed{2}

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Sir , Can you tell the motivation behind splitting the expression into a partial fraction and making it telescopic?

I mean how did you split it into a partial fraction?

Vilakshan Gupta - 3 years, 2 months ago

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Sums like this, as posted on Brilliant, nearly always work through being telescopic. It was obvious what the denominator ought to be, and it was simply a matter of playing with the numerator to get the right choice.

Mark Hennings - 3 years, 2 months ago
Naren Bhandari
Apr 4, 2018

∑ k = 1 ∞ 6 k ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) = ∑ k = 1 ∞ 2 k . 3 k ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) ∑ k = 1 ∞ 6 k ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) = ∑ k = 1 ∞ 2 k . ( ( 3 k + 1 − 2 k + 1 ) − 2 ( 3 k − 2 k ) ) ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) ∑ k = 1 ∞ 6 k ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) = ∑ k = 1 ∞ ( 2 k ( 3 k − 2 k ) − 2 k + 1 3 k + 1 − 2 k + 1 ) ∑ k = 1 ∞ 6 k ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) = 2 ∑ k = 1 ∞ ( 2 k − 1 ( 3 k − 2 k ) − 2 k − 2 3 k + 1 − 2 k + 1 ) \sum_{k=1}^{\infty} \dfrac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})} = \sum_{k=1}^{\infty} \dfrac{2^{k}.3^{k}}{(3^k-2^k)(3^{k+1}-2^{k+1})} \\ \sum_{k=1}^{\infty} \dfrac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})} = \sum_{k=1}^{\infty} \dfrac{2^{k}.((3^{k+1}-2^{k+1} )- 2(3^k -2^k))}{(3^k-2^k)(3^{k+1}-2^{k+1})} \\ \sum_{k=1}^{\infty} \dfrac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})} = \sum_{k=1}^{\infty} \left(\dfrac{2^k }{(3^k-2^k)} - \dfrac{2^{k+1}}{3^{k+1}-2^{k+1}}\right) \\ \sum_{k=1}^{\infty} \dfrac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})} = 2\sum_{k=1}^{\infty} \left(\dfrac{2^{k-1} }{(3^{k}-2^{k})} - \dfrac{2^{k-2}}{3^{k+1}-2^{k+1}}\right)

is telescoping series, hence the sum is 2 \boxed{2} .

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Alexander Becker
Apr 4, 2018

Source: Putnam 1984.

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