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Algebra

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26 Jan 2010 14:43:53 IST

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∞ 6^k ∑ -------------

None

∞ 6^k ∑ ------------------------------- = K=1 (3 ^k -2 ^k)(3^( k +1) –2 ^(k +1)

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Hari Shankar

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Joined: 28 Feb 2007

Posts: 2173

29 Jan 2010 14:57:54 IST

Like 3 people liked this

OMG! You have badly mangled this one up. I recognized it only because I had come across it earlier.

You are looking for:

$\sum_{k=1}^{\infty} \frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}$

Verify that you can write this as

$\sum_{k=1}^{\infty}\left[ \frac{3^k}{(3^k-2^k)} - \frac{3^{k+1}}{(3^{k+1}-2^{k+1})} \right]$

which is a telescopic summation. I am sure this can completed now

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