# summation

• Mar 8th 2010, 11:56 PM
hooke
summation
$\sum^{n}_{r=1}(\lg \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\lg 2^r+\lg (r+1)-\lg r]$

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

Am i correct , or it can be further simplified ? Thanks .
• Mar 9th 2010, 04:30 AM
tonio
Quote:

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

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

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

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

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

Tonio
• Mar 9th 2010, 04:33 AM
Hello hooke
Quote:

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

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

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

$\sum^{n}_{r=1}[\lg (r+1)-\lg r]=\lg(n+1)$
$\sum^{n}_{r=1}[\lg 2^r]=\lg 2\sum^{n}_{r=1}[r]$
$=\tfrac12n(n+1)\lg2$