Find:∑k=1∞6k(3k+1−2k+1)(3k−2k)

Question

ImpudgeIntemnect · Accepted Answer

That denominator should suggest the possibility of splitting the general term into partial fractions and getting a telescoping series of the form

\sum_{k \geq 1} (\frac{A_{k}}{3^{k} - 2^{k}} - \frac{A_{k + 1}}{3^{k + 1} - 2^{k + 1}})

where

A_{k}

very likely depends on k. Note that if this works, the sum of the series will be

\frac{A_{1}}{3^{1} - 2^{1}} = A_{1}

Now

\frac{A_{k}}{3^{k} - 2^{k}} - \frac{A_{k + 1}}{3^{k + 1} - 2^{k + 1}} = \frac{3^{k + 1} A_{k} - 3^{k} A_{k + 1} - 2^{k + 1} A_{k} + 2^{k} A_{k + 1}}{(3^{k} - 2^{k}) (3^{k + 1} - 2^{k + 1}}}

so you want to choose

A_{k}

and

A_{k + 1}

so that

3^{k + 1} A_{k} - 3^{k} A_{k + 1} - 2^{k + 1} A_{k} + 2^{k} A_{k + 1} = 6^{k}

The obvious things to try are

A_{k} = 2^{k}

, which makes the last two terms cancel out to leave

3^{k + 1} 2^{k} - 3^{k} 2^{k + 1} = 6^{k} (3 - 2) = 6^{k}

, and

A_{k} = 3^{k}

, which makes the first two terms cancel out and leaves

6^{k} (3 - 2) = 6^{k}

; both work.
However, summing

\sum_{k \geq 1} (\frac{A_{k}}{3^{k} - 2^{k}} - \frac{A_{k + 1}}{3^{k + 1} - 2^{k + 1}})

(1)
to

\frac{A_{1}}{3^{1} - 2^{1}} = A_{1}

is valid only if

lim_{k \to \infty} \frac{A_{k}}{3^{k} - 2^{k}} = 0

since the n-th partial sum of (1) is

A_{1} - \frac{A_{n + 1}}{3^{n + 1} - 2^{n + 1}}

Checking the two possibilities, we see that

lim_{k \to \infty} 2^{}

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