數學基礎之數列求和
阿新 • • 發佈:2021-12-12
錯位相減
證明等比數列求和 : \(\sum_{k=0}^{n}ap^{k} = a\frac{1-p^{n+1}}{1-p}\)
證明
\[\begin{aligned} S_n &=\sum_{k=0}^{n}ap^{k} = a\sum_{k=0}^{n}p^k\\ pS_n &= a\sum_{k=0}^{n}p^{k+1} = a\sum_{k=1}^{n+1}p^{k}\\ (1-p)S_n &= a(1 - p^{n+1})\\ S_n &= a\frac{1 - p^{n+1}}{1-p} \end{aligned} \]更一般的, 化簡 : \(\sum_{k=0}^{n}kp^k\)
裂項相消
化簡 \(S_n = \sum_{k=1}^{n}(ak + b)q^k\)
設 \((ak+b)q^k = f(k + 1)q^{k+1} - f(k)q^{k} , \quad f(k) = Ak + B\)
\[\begin{aligned} &(ak+b)q^k = (qf(k + 1) - f(k))q^{k}\\ &ak+b = qf(k+1) - f(k) = qA(k+1) + qB - Ak - B\\ &ak+b = qf(k+1) - f(k) = (q-1)Ak + qA + (q - 1)B\\ &A = \frac{a}{q-1}, B = \frac{b - qA}{q+1} = \frac{b - q\frac{a}{q-1}}{q-1}\\ \end{aligned} \]\(f(x)\)