WakeUp wrote: Let $P_k(x)=1+x+x^2+\ldots +x^{k-1}$. Show that \[ \sum_{k=1}^n \binom{n}{k} P_k(x)=2^{n-1} P_n \left( \frac{x+1}{2} \right) \] for every real number $x$ and every positive integer $n$. $P_k(x)=\frac{x^k-1}{x-1}$ $\forall x\ne 1$ $\sum_{k=1}^n\binom nkP_k(x)=\frac 1{x-1}\left(\sum_{k=1}^n\binom nkx^k-\sum_{k=1}^n\binom nk\right)$ $=\frac 1{x-1}\left((x+1)^n-1-(2^n-1)\right)$ $=2^n\frac 1{x-1}\left(\left(\frac{x+1}2\right)^n-1\right)$ $=2^{n-1}\frac {\left(\frac{x+1}2\right)^n-1}{\frac{x+1}2-1}$ $=2^{n-1}P_n\left(\frac{x+1}2\right)$ And this equality, true $\forall x\ne 1$ is still true for $x=1$ (direct verification or continuity, as you prefer). Q.E.D.