数学基础之数列求和
错位相减
证明等比数列求和 : \(\sum_{k=0}^{n}ap^{k} = a\frac{1-p^{n+1}}{1-p}\)
证明
\[\begin{aligned}
S_n &=\sum_{k=0}^{n}ap^{k} = a\sum_{k=0}^{n}p^k\\
pS_n &= a\sum_{k=0}^{n}p^{k+1} = a\sum_{k=1}^{n+1}p^{k}\\
(1-p)S_n &= a(1 - p^{n+1})\\
S_n &= a\frac{1 - p^{n+1}}{1-p}
\end{aligned}
\]
更一般的, 化简 : \(\sum_{k=0}^{n}kp^k\)
\[\begin{aligned}
S_n &= \sum_{k=0}^{n}kp^k\\
pS_n &= \sum_{k=0}^{n}kp^{k+1} = \sum_{k=1}^{n + 1}(k-1)p^{k}\\
(1-p)S_n &= \sum_{k=0}^{n}kp^k - \sum_{k=1}^{n + 1}(k-1)p^{k}\\
(1-p)S_n &= \sum_{k=1}^{n}kp^k - np^{n + 1} - \sum_{k=1}^{n}(k-1)p^{k}\\
(1-p)S_n &= \sum_{k=1}^{n}p^k - np^{n + 1}\\
(1-p)S_n &= p\frac{1-p^n}{1-p} - np^{n + 1}\\
S_n &= \frac{1}{1-p}(p\frac{1-p^n}{1-p} - np^{n + 1})\\
\end{aligned}
\]
裂项相消
化简 \(S_n = \sum_{k=1}^{n}(ak + b)q^k\)
设 \((ak+b)q^k = f(k + 1)q^{k+1} - f(k)q^{k} , \quad f(k) = Ak + B\)
\[\begin{aligned}
&(ak+b)q^k = (qf(k + 1) - f(k))q^{k}\\
&ak+b = qf(k+1) - f(k) = qA(k+1) + qB - Ak - B\\
&ak+b = qf(k+1) - f(k) = (q-1)Ak + qA + (q - 1)B\\
&A = \frac{a}{q-1}, B = \frac{b - qA}{q+1} = \frac{b - q\frac{a}{q-1}}{q-1}\\
\end{aligned}
\]
\(f(x)\) 已知, 所以
\[\sum_{k=1}^{n}a_k = f(n+1)q^{n + 1} - f(1)q
\]
