把差分算子作用在和式两边，可见若差分算子表达式的逆可以计算出来的话，和式便可求解。
S ( n ) = ∑ k = 1 n p ( k ) ⇒ Δ S ( n ) = p ( n + 1 ) ⇒ S ( n ) = Δ − 1 p ( n + 1 ) {\displaystyle S(n)=\sum _{k=1}^{n}p(k)\Rightarrow \Delta S(n)=p(n+1)\Rightarrow S(n)=\Delta ^{-1}p(n+1)}
但如裂项法那样，要求 x i = Δ y i {\displaystyle x_{i}=\Delta y_{i}} 的话，通常没有那么顺利。
∑ k = 1 n p ( k ) q k − 1 = f ( n ) q n − f ( 0 ) {\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0)} f ( n ) = p ( n ) q − 1 + 1 ( q − 1 ) 2 ∑ k = 1 m ( − 1 ) k q k − 1 ( q − 1 ) k − 1 Δ k ( p ( n ) ) = 1 q − 1 ∑ k = 0 m ( − q q − 1 ) k Δ k p ( n + 1 ) {\displaystyle f(n)={\frac {p(n)}{q-1}}+{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}{\frac {(-1)^{k}q^{k-1}}{(q-1)^{k-1}}}\Delta ^{k}(p(n))={\frac {1}{q-1}}\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}p(n+1)} [1] 其中 Δ p ( n ) = p ( n + 1 ) − p ( n ) , m = d e g ( p ) {\displaystyle \Delta p(n)=p(n+1)-p(n),m=deg(p)}
∑ k = 1 n p ( k ) q k − 1 = f ( n ) q n − f ( 0 ) {\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0)}
f ( n ) = p ( n ) q − 1 + 1 ( q − 1 ) 2 ∑ k = 1 m ( − 1 ) k q k − 1 ( q − 1 ) k − 1 Δ k ( p ( n ) ) = 1 q − 1 ∑ k = 0 m ( − q q − 1 ) k Δ k p ( n + 1 ) {\displaystyle f(n)={\frac {p(n)}{q-1}}+{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}{\frac {(-1)^{k}q^{k-1}}{(q-1)^{k-1}}}\Delta ^{k}(p(n))={\frac {1}{q-1}}\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}p(n+1)} [1]
其中 Δ p ( n ) = p ( n + 1 ) − p ( n ) , m = d e g ( p ) {\displaystyle \Delta p(n)=p(n+1)-p(n),m=deg(p)}
这是显式地将和式写成差分算子表达式的一个例子。
∑ 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}}} 设 ∑ k = 1 n p ( k ) q k − 1 = f ( n ) q n + C {\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}+C} 当 m = d e g ( p ) = 1 {\displaystyle m=deg(p)=1} 时， d e g ( f ) = m = 1 {\displaystyle deg(f)=m=1} 两边逐项求导： ∑ k = 1 n ( k − 1 ) p ( k ) q k − 2 = [ d f ( n ) d q + n f ( n ) q − 1 ] q n + d C d q {\displaystyle \sum _{k=1}^{n}(k-1)p(k)q^{k-2}=[{\frac {df(n)}{dq}}+nf(n)q^{-1}]q^{n}+{\frac {dC}{dq}}} 可见两边多项式的阶数同时增加1 ∑ k = 0 n p ( k ) q k − 1 = f ( n ) q n + C + p ( 0 ) q − 1 {\displaystyle \sum _{k=0}^{n}p(k)q^{k-1}=f(n)q^{n}+C+p(0)q^{-1}} 代入 n = 0 {\displaystyle n=0} 得到 p ( 0 ) q − 1 = f ( 0 ) + C + p ( 0 ) q − 1 ⇒ C = − f ( 0 ) {\displaystyle p(0)q^{-1}=f(0)+C+p(0)q^{-1}\Rightarrow C=-f(0)}
对 ∑ k = 1 n p ( k ) q k − 1 = f ( n ) q n − f ( 0 ) {\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0)} 两边差分 p ( n + 1 ) q n = f ( n + 1 ) q n + 1 − f ( n ) q n {\displaystyle p(n+1)q^{n}=f(n+1)q^{n+1}-f(n)q^{n}} p ( n + 1 ) = q f ( n + 1 ) − f ( n ) {\displaystyle p(n+1)=qf(n+1)-f(n)} p ( n + 1 ) = p ( n ) + Δ p ( n ) = ( I + Δ ) p ( n ) {\displaystyle p(n+1)=p(n)+\Delta p(n)=(I+\Delta )p(n)} ( I + Δ ) p ( n ) = q ( I + Δ ) f ( n ) − f ( n ) = [ ( q − 1 ) I + q Δ ] f ( n ) {\displaystyle (I+\Delta )p(n)=q(I+\Delta )f(n)-f(n)=[(q-1)I+q\Delta ]f(n)} f ( n ) = I + Δ ( q − 1 ) I + q Δ p ( n ) = 1 ( q − 1 ) I + q Δ p ( n + 1 ) {\displaystyle f(n)={\frac {I+\Delta }{(q-1)I+q\Delta }}p(n)={\frac {1}{(q-1)I+q\Delta }}p(n+1)}
由于f,p都是m次多项式，以下可进行满足条件 Δ m + 1 = 0 {\displaystyle \Delta ^{m+1}=0} 的求逆运算：
I + Δ ( q − 1 ) I + q Δ = I + Δ q − 1 ∑ k = 0 m ( − q q − 1 ) k Δ k {\displaystyle {\frac {I+\Delta }{(q-1)I+q\Delta }}={\frac {I+\Delta }{q-1}}\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}} = 1 q − 1 [ ∑ k = 0 m ( − q q − 1 ) k Δ k + ∑ k = 0 m ( − q q − 1 ) k Δ k + 1 ] = 1 q − 1 [ ∑ k = 0 m ( − q q − 1 ) k Δ k + ∑ k = 1 m ( − q q − 1 ) k − 1 Δ k ] {\displaystyle ={\frac {1}{q-1}}[\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}+\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k+1}]={\frac {1}{q-1}}[\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}+\sum _{k=1}^{m}({\frac {-q}{q-1}})^{k-1}\Delta ^{k}]} = 1 q − 1 I + 1 q − 1 ∑ k = 1 m [ ( − q q − 1 ) k + ( − q q − 1 ) k − 1 ] Δ k = 1 q − 1 I − 1 ( q − 1 ) 2 ∑ k = 1 m ( − q q − 1 ) k − 1 Δ k {\displaystyle ={\frac {1}{q-1}}I+{\frac {1}{q-1}}\sum _{k=1}^{m}[({\frac {-q}{q-1}})^{k}+({\frac {-q}{q-1}})^{k-1}]\Delta ^{k}={\frac {1}{q-1}}I-{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}({\frac {-q}{q-1}})^{k-1}\Delta ^{k}} = 1 q − 1 I + 1 ( q − 1 ) 2 ∑ k = 1 m ( − 1 ) k q k − 1 ( q − 1 ) k − 1 Δ k {\displaystyle ={\frac {1}{q-1}}I+{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}{\frac {(-1)^{k}q^{k-1}}{(q-1)^{k-1}}}\Delta ^{k}} 最后把算子表达式作用到p上得证。
计算各阶差分： p ( n ) = a + ( n − 1 ) d , Δ p ( n ) = d {\displaystyle p(n)=a+(n-1)d,\Delta p(n)=d}
f ( n ) = a + ( n − 1 ) d r − 1 − d ( r − 1 ) 2 {\displaystyle \displaystyle f(n)={\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}}
∑ k = 1 n [ a + ( k − 1 ) d ] r k − 1 = [ a + ( n − 1 ) d r − 1 − d ( r − 1 ) 2 ] r n − [ a − d r − 1 − d ( r − 1 ) 2 ] {\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=[{\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]}
计算各阶差分： p ( n ) = A + B n + C n 2 , Δ p ( n ) = B + C ( 2 n + 1 ) = B + C + 2 C n , Δ 2 p ( n ) = 2 C {\displaystyle p(n)=A+Bn+Cn^{2},\Delta p(n)=B+C(2n+1)=B+C+2Cn,\Delta ^{2}p(n)=2C}
f ( n ) = A + B n + C n 2 q − 1 − B + C + 2 C n ( q − 1 ) 2 + 2 C ( q − 1 ) 3 q {\displaystyle \displaystyle f(n)={\frac {A+Bn+Cn^{2}}{q-1}}-{\frac {B+C+2Cn}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}q}
∑ k = 1 n p ( k ) q k − 1 = [ A + B n + C n 2 q − 1 − B + C + 2 C n ( q − 1 ) 2 + 2 C ( q − 1 ) 3 q ] q n − [ A q − 1 − B + C ( q − 1 ) 2 + 2 C ( q − 1 ) 3 q ] {\displaystyle \displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=[{\frac {A+Bn+Cn^{2}}{q-1}}-{\frac {B+C+2Cn}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}q]q^{n}-[{\frac {A}{q-1}}-{\frac {B+C}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}q]}
∑ k = 1 ∞ p ( k ) q k − 1 = p ( 0 ) 1 − q + 1 ( 1 − q ) 2 ∑ k = 1 m q k − 1 ( 1 − q ) k − 1 Δ k ( p ( 0 ) ) , | q | < 1 {\displaystyle \sum _{k=1}^{\infty }p(k)q^{k-1}={\frac {p(0)}{1-q}}+{\frac {1}{(1-q)^{2}}}\sum _{k=1}^{m}{\frac {q^{k-1}}{(1-q)^{k-1}}}\Delta ^{k}(p(0)),~|q|<1}
∑ k = 1 ∞ ( A + B k + C k 2 ) q k − 1 = A 1 − q + B + C ( 1 − q ) 2 + 2 C ( 1 − q ) 3 q {\displaystyle \displaystyle \sum _{k=1}^{\infty }(A+Bk+Ck^{2})q^{k-1}={\frac {A}{1-q}}+{\frac {B+C}{(1-q)^{2}}}+{\frac {2C}{(1-q)^{3}}}q}
的话，通常没有那么顺利。
[1]
![{\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=[{\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc72cab8df2afc68e6f8c06531dca45d8ad47d68)
![{\displaystyle \displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=[{\frac {A+Bn+Cn^{2}}{q-1}}-{\frac {B+C+2Cn}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}q]q^{n}-[{\frac {A}{q-1}}-{\frac {B+C}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}q]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe3d21a17e7f279d1573dea00d5c832f56146421)
![{\displaystyle \sum _{k=1}^{n}(k-1)p(k)q^{k-2}=[{\frac {df(n)}{dq}}+nf(n)q^{-1}]q^{n}+{\frac {dC}{dq}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94ff7b62da0e9b12bc5f4c53833ef111719f50f6)
![{\displaystyle (I+\Delta )p(n)=q(I+\Delta )f(n)-f(n)=[(q-1)I+q\Delta ]f(n)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b47f2425ac73cbefaed28e1494fe7916d834c9d)
![{\displaystyle ={\frac {1}{q-1}}[\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}+\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k+1}]={\frac {1}{q-1}}[\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}+\sum _{k=1}^{m}({\frac {-q}{q-1}})^{k-1}\Delta ^{k}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/923e5b28c1d4aefcb03b521620ab098bd4c99ada)
![{\displaystyle ={\frac {1}{q-1}}I+{\frac {1}{q-1}}\sum _{k=1}^{m}[({\frac {-q}{q-1}})^{k}+({\frac {-q}{q-1}})^{k-1}]\Delta ^{k}={\frac {1}{q-1}}I-{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}({\frac {-q}{q-1}})^{k-1}\Delta ^{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a983df4fd109ccff6259aea3756dd6f51d6b080)
