$k> 0$ 。当 $k$为奇数时，
\begin{aligned}
\sum_{i = 0}^{k} \binom{n}{i} &= [\binom{n}{0} + \binom{n}{1} ] + [\binom{n}{2} + \binom{n}{3} ] + \dots + [\binom{n}{k-1} + \binom{n}{k} ] \\
&= \binom{n+1}{1} + \binom{n+1}{3} + \dots + \binom{n+1}{k}
\end{aligned}
又有
\begin{aligned}
\sum_{i = 0}^{k} \binom{n}{i} &= \binom{n}{0} + [\binom{n}{1} + \binom{n}{2}] + [\binom{n}{3} + \binom{n}{4}] + \dots + [\binom{n}{k-2} + \binom{n}{k-1}] + \binom{n}{k} \\
&= \binom{n}{0} + \binom{n+1}{2} + \binom{n+1}{4} + \dots + \binom{n+1}{k-1} + \binom{n}{k} \\
&= \binom{n+1}{0} + \binom{n+1}{2} + \binom{n+1}{4} + \dots + \binom{n+1}{k-1} + \binom{n}{k}
\end{aligned}
两式相加得
\begin{aligned}
2 \sum_{i = 0}^{k} \binom{n}{i} = \binom{n}{k} + \sum_{i=0}^{k} \binom{n+1}{i}
\end{aligned}
当 $k$ 为偶数时，经过类似的推导，可得
\begin{aligned}
2 \sum_{i = 0}^{k} \binom{n}{i} = \binom{n}{k} + \sum_{i=0}^{k} \binom{n+1}{i}
\end{aligned}
容易验证，当 $k = 0$ 时，上式仍成立。于是，对任意 $0 \le k \le n$ 有
\begin{aligned}
2 \sum_{i = 0}^{k} \binom{n}{i} = \binom{n}{k} + \sum_{i=0}^{k} \binom{n+1}{i}
\end{aligned}
亦即
\begin{equation}
\sum_{i=0}^{k} \binom{n+1}{i} = 2 \sum_{i = 0}^{k} \binom{n}{i} - \binom{n}{k} \label{E:1}
\end{equation}
为了简便，将 $ \sum_{i=0}^{k} \binom{n}{i}$ 记做 $S(n, k)$ 。反复使用 \eqref{E:1} 式，可以得到
\begin{aligned}
S(n, k) = 2^{n-k} S(k, k) - 2^{n - k - 1} \binom{k}{k} - 2^{n - k - 2} \binom{k+1}{k} - \dots - 2^{1} \binom{n-2}{k} - 2^{0} \binom{n-1}{k}
\end{aligned}