代數/本書課文/求和/逐項求導法

對已知的和式逐項求導，如$S(n,q)=\sum _{k=1}^{n}p(k)q^{k-1}$，得到未知的和式。

Template:ExampleRobox $\sum _{k=1}^n{q}^{k-1}=\frac{q^n-1}{q-1}$
$\sum _{k=1}^{n}k{q}^{k-1}=\frac{d}{dq}(\frac{q^{n+1}-q}{q-1})=\frac{(n+1)q^n-1}{q-1}-\frac{q^{n+1}-q}{(q-1{)}^2}=\frac{n{q}^{n+1}-(n+1)q^n+1}{(q-1{)}^2}$
$\sum _{k=1}^n{k}^2{q}^{k-1}=\frac{d}{dq}[\frac{n{q}^{n+2}-(n+1)q^{n+1}+q}{(q-1{)}^2}]=\frac{n(n+2)q^{n+1}-(n+1{)}^2{q}^n+1}{(q-1{)}^2}-2\frac{n{q}^{n+2}-(n+1)q^{n+1}+q}{(q-1{)}^3}$
$=\frac{n^2{q}^{n+2}-(2n^2+2n-1)q^{n+1}+(n+1{)}^2{q}^n-q-1}{(q-1{)}^3}$
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Template:ExampleRobox $\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots$
兩邊逐項求導，得到：$\sum _{n=1}^{\mathrm{\infty }}n{x}^{n-1}=\frac{1}{(1-x{)}^2},|x|<1.$
求m次導,得到:$\sum _{n=1}^{\mathrm{\infty }}{C}_n^m{x}^{n-m}=\frac{m!}{(1-x{)}^{m+1}},|x|<1.$ Template:Robox/Close