Evaluate the expression for $\alpha \in \mathbf{R}$: $\sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}C_{n}^k$

Question

I've tried using binomial theorem for $(1+\alpha)^{n}$, but the expanding denominator is a problem. The same case for using property $C_{n}^k$ = $C_{n}^{n-k}$ and summing opposite components (e.g. $\frac{a^3}{3}C_{n}^2 + \frac{a^{n-1}}{n-1}C_{n}^{n-2} = \frac{a^3}{3}C_{n}^2 + \frac{a^{n-1}}{n-1}C_{n}^{2} = C_{n}^{2}(\frac{a^3}{3} + \frac{a^{n-1}}{n-1})$ etc.)

Answer 1

Begin by writing that for all $x$, $$ (x + 1)^n = \sum_{k = 0}^n C_n^kx^k, $$ and then compute the integral of this expression from $0$ to some real number $a$ and use the liearity of the integral, $$ \int_0^a (x + 1)^n \, dx = \int_0^a \sum_{k = 0}^n C_n^kx^k \, dx = \sum_{k = 0}^n C_n^k \int_0^a x^k \, dx = \sum_{k = 0}^n C_n^k\frac{a^{k + 1}}{k + 1}. $$ Finally, you sum equals, $$ \sum_{k = 0}^n C_n^k\frac{a^{k + 1}}{k + 1} = \int_0^a (x + 1)^n \, dx = \frac{(a + 1)^{n + 1} - 1}{n + 1}. $$ You can also use the fact that $\frac{1}{k + 1}C_n^k = \frac{1}{n + 1}C_{n + 1}^{k + 1}$ and use directly the binomial formula.