an easy binomial

Using the binomial theorem as follows:

$\begin{aligned} \sum_{k=0}^n \binom nk u^k & = (1+u)^n & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }u \\ \sum_{k=0}^n k \binom nk u^{k-1} & = n(1+u)^{n-1} & \small \color{#3D99F6} \text{Multiply both sides by }u \\ \sum_{k=0}^n k \binom nk u^k & = nu(1+u)^{n-1} & \small \color{#3D99F6} \text{Integrate both sides w.r.t. }u \\ \sum_{k=0}^n \frac k{k+1} \binom nk u^{k+1} & = \color{#3D99F6} u(1+u)^n - \int (1+u)^n du & \small \color{#3D99F6} \text{By integration by parts.} \\ & = u(1+u)^n - \frac {(1+u)^{n+1}}{n+1} + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = u(1+u)^n - \frac {(1+u)^{n+1}}{n+1} + \color{#3D99F6} \frac 1{n+1} & \small \color{#3D99F6} \text{when }u=0 \implies C = \frac 1{n+1} \\ & = \frac {(1+u)^n(u(n+1) - (1+u)) + 1}{n+1} \\ & = \frac {(1+u)^n(nu - 1) + 1}{n+1} & \small \color{#3D99F6} \text{Put }u=1 \\ \sum_{k=0}^n \frac k{k+1} \binom nk & = \frac {2^n(n - 1) + 1}{n+1} & \small \color{#3D99F6} \text{Put }n=100 \\ \sum_{k=0}^{100} \frac k{k+1} \binom {100}k & = \frac {2^{100}(99) + 1}{101} \end{aligned}$

Therefore, $xyz = 99 \times 1 \times 101 = \boxed{9999}$ .

N(x+1)^(n-1)+(x+1)^n×x+1