d+e) Ta có $S_4=\sum_{k=0}^{n}C^k_n.2^n=(1+2)^n=3^n$. Mặt khác
Đặt $S_6=C^1_n2^1+C^3_n2^3+C^5_n2^5+\ldots$ thì
$S_4=S_5+S_6$
$(-1)^n=(1-2)^n=\sum_{k=0}^{n}C^k_n2^k.(-1)^{k}=S_5-S_6$.
Vậy $S_5=\frac{1}{2}\left ( S_4+(-1)^n \right )=\frac{1}{2}\left ( 3^n+(-1)^n \right ).$