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The following question is:

Show that $\sum\limits_{r = 1}^n {r(r + 2)} ={n \over 6}(n+1)(2n+7).$

Using this results, or otherwise, find, in terms of $n$, the sum of the series $$3\ln2+4\ln2^2+5\ln2^3+\dots +(n+2)\ln(2^n)$$ Express your answer in its simplest form.

ps: I have shown the result but having problem with the second part which is to find the sum of the series. I had no idea where to start.

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Hint

Since $\log(2^n)=n\log(2)$, then $$\sum_{i=1}^n (i+2)\log(2^i)=\sum_{i=1}^n i(i+2)\log(2)=\log(2) \sum_{i=1}^n i(i+2)$$

I am sure that you can take from here

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$$\sum_{r=1}^nr(r+2)=\sum_{r=1}^nr^2+2\sum_{r=1}^nr$$

See sum of first $n$ natural number $\dfrac{n(n+1)}2$

and How to get to the formula for the sum of squares of first n numbers?

Now, $$\ln2^r=r\ln2\implies (r+2)(\ln2^r)=r(r+2)\ln2$$

$$\implies\sum_{r=1}^n(r+2)(\ln2^r)=\ln2\sum_{r=1}^nr(r+2)=?$$

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  • $\begingroup$ You won AGAIN ! Cheers :) $\endgroup$ – Claude Leibovici Sep 20 '14 at 14:33

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