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 .
I agree with the final term of your answer:$\displaystyle \sum^{n}_{r=1}[\lg (r+1)-\lg r]=\lg(n+1)$
But the first term is simply an AP:$\displaystyle \sum^{n}_{r=1}[\lg 2^r]=\lg 2\sum^{n}_{r=1}[r]$$\displaystyle =\tfrac12n(n+1)\lg2$
Grandad