1. ## Finite Sum

Show that $\displaystyle \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}-1}{n+1}$.

2. Originally Posted by Wizard2010
Show that $\displaystyle \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}-1}{n+1}$.
What have you tried? Have you hit it with induction yet?

3. Originally Posted by Swlabr
What have you tried? Have you hit it with induction yet?
I've actually realised that it's pretty easy. I've shown it (not by induction). Thanks.

4. Originally Posted by Wizard2010
Show that $\displaystyle \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}-1}{n+1}$.
$\displaystyle \displaystyle (1+\alpha)^n = \sum_{k=0}^{n}\binom{n}{k}\alpha^k$, therefore $\displaystyle \displaystyle \int(1+\alpha)^n \;{d\alpha} = \int\sum_{k=0}^{n}\binom{n}{k}\alpha^k\;{d\alpha}.$

So $\displaystyle \displaystyle \frac{(1+\alpha)^{n+1}}{n+1} + C = \sum_{k=0}^n \frac{\alpha^{k+1}}{k+1}\binom{n}{k}.$ Put $\displaystyle \alpha = 0$, to get $\displaystyle C = -\frac{1}{n+1}$.

Thus $\displaystyle \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}}{n+1}-\frac{1}{n+1} = \frac{(\alpha+1)^{n+1}-1}{n+1}.$