Using the result $$\sum_{r=1}^nr(r+2)=\frac n6(n+1)(2n+7)$$ find, interms of $n$, the sum of the series
$$3\ln2+4\ln2^2+5\ln2^3+...+(n+2)\ln2^n$$ and express in its simplest form.
Where do I start please?
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$\begingroup$ What is a proof that $\sum_{r=1}^n r(r+2) = \frac{n}{6} (n+1)(2n+7)$? Does anyone have a link? $\endgroup$– Eric AuldAug 24, 2013 at 16:53
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2$\begingroup$ Well, $r(r+2)=(r+1)^2-1$. So it shouldn't be hard to prove. @EricAuld $\endgroup$– Thomas AndrewsAug 24, 2013 at 16:55
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2 Answers
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HINT:
As $\ln a^b=b\ln a,$
$$3\ln2+4\ln2^2+5\ln2^3+...+(n+2)\ln2^n=\sum_{1\le r\le n}(r+2)\ln 2^r=\ln 2\sum_{1\le r\le n}r(r+2)$$
answered Aug 24, 2013 at 16:49
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HINT:
Use the fact that: $$ \ln (a^x) = x\ln(a) $$
answered Aug 24, 2013 at 16:48