1. ## summation

$\displaystyle \sum^{n}_{r=1}(\lg \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\lg 2^r+\lg (r+1)-\lg r]$

i found the answer to be $\displaystyle (2^n-1)\lg 2 +\lg (n+1)$

Am i correct , or it can be further simplified ? Thanks .

2. Originally Posted by hooke
$\displaystyle \sum^{n}_{r=1}(\lg \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\lg 2^r+\lg (r+1)-\lg r]$

i found the answer to be $\displaystyle (2^n-1)\lg 2 +\lg (n+1)$

Am i correct , or it can be further simplified ? Thanks .

How did you get that? Assuming lg means log, I get:

$\displaystyle \sum^{n}_{r=1}(\log \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\log 2^r+\log (r+1)-\log r]$ $\displaystyle =\log 2\sum^n_{r=1}r+\log (n+1)=\frac{n(n+1)}{2}\,\log 2+\log (n+1)$

Tonio

3. Hello hooke
Originally Posted by hooke
$\displaystyle \sum^{n}_{r=1}(\lg \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\lg 2^r+\lg (r+1)-\lg r]$

i found the answer to be $\displaystyle (2^n-1)\lg 2 +\lg (n+1)$

Am i correct , or it can be further simplified ? Thanks .
$\displaystyle \sum^{n}_{r=1}[\lg (r+1)-\lg r]=\lg(n+1)$
$\displaystyle \sum^{n}_{r=1}[\lg 2^r]=\lg 2\sum^{n}_{r=1}[r]$
$\displaystyle =\tfrac12n(n+1)\lg2$