0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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 Auld
    Commented Aug 24, 2013 at 16:53
  • 2
    $\begingroup$ Well, $r(r+2)=(r+1)^2-1$. So it shouldn't be hard to prove. @EricAuld $\endgroup$ Commented Aug 24, 2013 at 16:55

2 Answers 2

1
$\begingroup$

HINT:

As $\ln a^b=b\ln a,$

$$3\ln2+4\ln2^2+5\ln2^3+...+(n+2)\ln2^n=\sum_{1\le r\le n}(r+2)\ln 2^r=\ln 2\sum_{1\le r\le n}r(r+2)$$

$\endgroup$
0
1
$\begingroup$

HINT:

Use the fact that: $$ \ln (a^x) = x\ln(a) $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .