對已知的和式逐項求導，如 S ( n , q ) = ∑ k = 1 n p ( k ) q k − 1 {\displaystyle S(n,q)=\sum _{k=1}^{n}p(k)q^{k-1}} ，得到未知的和式。
∑ k = 1 n q k − 1 = q n − 1 q − 1 {\displaystyle \sum _{k=1}^{n}q^{k-1}={\frac {q^{n}-1}{q-1}}} ∑ k = 1 n k q k − 1 = d d q ( q n + 1 − q q − 1 ) = ( n + 1 ) q n − 1 q − 1 − q n + 1 − q ( q − 1 ) 2 = n q n + 1 − ( n + 1 ) q n + 1 ( q − 1 ) 2 {\displaystyle \sum _{k=1}^{n}kq^{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 {nq^{n+1}-(n+1)q^{n}+1}{(q-1)^{2}}}} ∑ k = 1 n k 2 q k − 1 = d d q [ n q n + 2 − ( n + 1 ) q n + 1 + q ( q − 1 ) 2 ] = n ( n + 2 ) q n + 1 − ( n + 1 ) 2 q n + 1 ( q − 1 ) 2 − 2 n q n + 2 − ( n + 1 ) q n + 1 + q ( q − 1 ) 3 {\displaystyle \sum _{k=1}^{n}k^{2}q^{k-1}={\frac {d}{dq}}[{\frac {nq^{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 {nq^{n+2}-(n+1)q^{n+1}+q}{(q-1)^{3}}}} = n 2 q n + 2 − ( 2 n 2 + 2 n − 1 ) q n + 1 + ( n + 1 ) 2 q n − q − 1 ( q − 1 ) 3 {\displaystyle ={\frac {n^{2}q^{n+2}-(2n^{2}+2n-1)q^{n+1}+(n+1)^{2}q^{n}-q-1}{(q-1)^{3}}}}
1 1 − x = 1 + x + x 2 + x 3 + x 4 + ⋯ {\displaystyle {\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+\cdots } 兩邊逐項求導，得到： ∑ n = 1 ∞ n x n − 1 = 1 ( 1 − x ) 2 , | x | < 1. {\displaystyle \sum _{n=1}^{\infty }nx^{n-1}={\frac {1}{(1-x)^{2}}}\quad ,|x|<1.} 求m次導,得到: ∑ n = 1 ∞ C n m x n − m = m ! ( 1 − x ) m + 1 , | x | < 1. {\displaystyle \sum _{n=1}^{\infty }C_{n}^{m}x^{n-m}={\frac {m!}{(1-x)^{m+1}}}\quad ,|x|<1.}
，得到未知的和式。
![{\displaystyle \sum _{k=1}^{n}k^{2}q^{k-1}={\frac {d}{dq}}[{\frac {nq^{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 {nq^{n+2}-(n+1)q^{n+1}+q}{(q-1)^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c8db76b17a2d6386bd1e98af8c6445fb02bee63)
