在数学上，等差-等比数列（简称差比数列，英语：arithmetico-geometric sequence）是一个等差数列与一个等比数列相乘的积。 通项公式 等差-等比数列有如下通项公式：[1] [ a + ( n − 1 ) d ] r n − 1 {\displaystyle [a+(n-1)d]r^{n-1}} 其中 r {\displaystyle r} 是公比，而 r n − 1 {\displaystyle r^{n-1}} 的系数： [ a + ( n − 1 ) d ] {\displaystyle [a+(n-1)d]} 则是等差数列的项，其首项为 a {\displaystyle a} ，公差 d {\displaystyle d} 。 等差-等比数列的求和公式 等差-等比级数有如下形式； ∑ k = 1 n [ a + ( k − 1 ) d ] r k − 1 = a + ( a + d ) r + ( a + 2 d ) r 2 + ⋯ + [ a + ( n − 1 ) d ] r n − 1 {\displaystyle \sum _{k=1}^{n}\left[a+(k-1)d\right]r^{k-1}=a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}} 其前n项之和为； S n = ∑ k = 1 n [ a + ( k − 1 ) d ] r k − 1 = a 1 − r − [ a + ( n − 1 ) d ] r n 1 − r + d r ( 1 − r n − 1 ) ( 1 − r ) 2 . {\displaystyle S_{n}=\sum _{k=1}^{n}\left[a+(k-1)d\right]r^{k-1}={\frac {a}{1-r}}-{\frac {[a+(n-1)d]r^{n}}{1-r}}+{\frac {dr(1-r^{n-1})}{(1-r)^{2}}}.} 错位相减法 由此级数开始：[1][2] S n = a + ( a + d ) r + ( a + 2 d ) r 2 + ⋯ + [ a + ( n − 1 ) d ] r n − 1 {\displaystyle S_{n}=a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}} 将Sn乘以r， r S n = a r + ( a + d ) r 2 + ( a + 2 d ) r 3 + ⋯ + [ a + ( n − 1 ) d ] r n {\displaystyle rS_{n}=ar+(a+d)r^{2}+(a+2d)r^{3}+\cdots +[a+(n-1)d]r^{n}} Sn减去rSn， S n ( 1 − r ) = { a + ( a + d ) r + ( a + 2 d ) r 2 + ⋯ + [ a + ( n − 1 ) d ] r n − 1 } − { a r + ( a + d ) r 2 + ( a + 2 d ) r 3 + ⋯ + [ a + ( n − 1 ) d ] r n } = a + [ d r + d r 2 + ⋯ + d r n − 1 ] − [ a + ( n − 1 ) d ] r n = a + [ d r ( 1 − r n − 1 ) 1 − r ] − [ a + ( n − 1 ) d ] r n {\displaystyle {\begin{aligned}S_{n}(1-r)&=&\left\{a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}\right\}\\&&-\left\{ar+(a+d)r^{2}+(a+2d)r^{3}+\cdots +[a+(n-1)d]r^{n}\right\}\\&=&a+\left[dr+dr^{2}+\cdots +dr^{n-1}\right]-[a+(n-1)d]r^{n}\\&=&a+\left[{\frac {dr(1-r^{n-1})}{1-r}}\right]-[a+(n-1)d]r^{n}\end{aligned}}} 在中间的项中使用等比数列的求和公式。最后左右两边同除以(1 − r)，得到最终结果。 逐项求导 ∑ k = 1 n r k − 1 = r n − 1 r − 1 {\displaystyle \displaystyle \sum _{k=1}^{n}r^{k-1}={\frac {r^{n}-1}{r-1}}} 对等比数列和两边求导：[3] ∑ k = 1 n ( k − 1 ) r k − 2 = n r n − 1 r − 1 − r n − 1 ( r − 1 ) 2 {\displaystyle \displaystyle \sum _{k=1}^{n}(k-1)r^{k-2}={\frac {nr^{n-1}}{r-1}}-{\frac {r^{n}-1}{(r-1)^{2}}}} ∑ k = 1 n [ a + ( k − 1 ) d ] r k − 1 = a r n − 1 r − 1 + d r [ n r n − 1 r − 1 − r n − 1 ( r − 1 ) 2 ] {\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=a{\frac {r^{n}-1}{r-1}}+dr[{\frac {nr^{n-1}}{r-1}}-{\frac {r^{n}-1}{(r-1)^{2}}}]} 裂项法 待定系数s,t使得等差-等比数列可以裂项：[4] [ a + ( k − 1 ) d ] r k − 1 = ( s k + t ) r k − [ s ( k − 1 ) + t ] r k − 1 {\displaystyle [a+(k-1)d]r^{k-1}=(sk+t)r^{k}-[s(k-1)+t]r^{k-1}} 用裂项法可以求出数列和： ∑ k = 1 n [ a + ( k − 1 ) d ] r k − 1 = ( s n + t ) r n − t {\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=(sn+t)r^{n}-t} 求出待定系数s,t关于a,d,r的表达式： d k + a − d = s ( r − 1 ) k + ( r − 1 ) t + s {\displaystyle dk+a-d=s(r-1)k+(r-1)t+s} s = d r − 1 , t = a − d r − 1 − d ( r − 1 ) 2 {\displaystyle \displaystyle s={\frac {d}{r-1}},t={\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}} ∑ k = 1 n [ a + ( k − 1 ) d ] r k − 1 = [ d r − 1 n + a − 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 {d}{r-1}}n+{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]} 差分算子公式 ∑ 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 \displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0),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)} [5] 其中 Δ p ( n ) = p ( n + 1 ) − p ( n ) {\displaystyle \Delta p(n)=p(n+1)-p(n)} 求出各阶差分： 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}}}]} 无穷级数 如果 − 1 < r < 1 {\displaystyle -1<r<1} ，那么其无穷级数为[1] lim n → ∞ S n = a 1 − r + d r ( 1 − r ) 2 {\displaystyle \lim _{n\to \infty }S_{n}={\frac {a}{1-r}}+{\frac {dr}{(1-r)^{2}}}} 如果 r {\displaystyle r} 在上述范围之外，则该级数不是发散级数就是交错级数。 参见 序列 参考文献Loading content...Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.
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![{\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}}}]}](//wikimedia.org/api/rest_v1/media/math/render/svg/bc72cab8df2afc68e6f8c06531dca45d8ad47d68)
![{\displaystyle [a+(n-1)d]r^{n-1}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/0c98b4f9faa620ac0b0b6f32e09cae6ee75216b9)
![{\displaystyle [a+(n-1)d]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/2fdade0a828ef6c5fafd1ad31e1333e1f8fdf4a5)
![{\displaystyle \sum _{k=1}^{n}\left[a+(k-1)d\right]r^{k-1}=a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1cb778ef20d37ad57e0f9188ad365ad17de866d8)
![{\displaystyle S_{n}=\sum _{k=1}^{n}\left[a+(k-1)d\right]r^{k-1}={\frac {a}{1-r}}-{\frac {[a+(n-1)d]r^{n}}{1-r}}+{\frac {dr(1-r^{n-1})}{(1-r)^{2}}}.}](http://wikimedia.org/api/rest_v1/media/math/render/svg/5d467c0294256a6823d2885db162224ba0951072)
![{\displaystyle S_{n}=a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/2ce10c2a23b8dbc6db1b31c312f470fc04194032)
![{\displaystyle rS_{n}=ar+(a+d)r^{2}+(a+2d)r^{3}+\cdots +[a+(n-1)d]r^{n}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/f7bace018677fc5840cfae99964ec8b5358b693e)
![{\displaystyle {\begin{aligned}S_{n}(1-r)&=&\left\{a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}\right\}\\&&-\left\{ar+(a+d)r^{2}+(a+2d)r^{3}+\cdots +[a+(n-1)d]r^{n}\right\}\\&=&a+\left[dr+dr^{2}+\cdots +dr^{n-1}\right]-[a+(n-1)d]r^{n}\\&=&a+\left[{\frac {dr(1-r^{n-1})}{1-r}}\right]-[a+(n-1)d]r^{n}\end{aligned}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/6d158643292018dc1ad6e84c188ef0ba85ab7e5b)
![{\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=a{\frac {r^{n}-1}{r-1}}+dr[{\frac {nr^{n-1}}{r-1}}-{\frac {r^{n}-1}{(r-1)^{2}}}]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/249b87c7b48b45715959865d069306b1caefef40)
![{\displaystyle [a+(k-1)d]r^{k-1}=(sk+t)r^{k}-[s(k-1)+t]r^{k-1}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/bab1a711d703256fc78e31218ea8e556dbc4ade2)
![{\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=(sn+t)r^{n}-t}](http://wikimedia.org/api/rest_v1/media/math/render/svg/40c9cba3e04c8d4bb3a21f2ff350c85657556626)
![{\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=[{\frac {d}{r-1}}n+{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/c45bcfeb0b578847d2748f77336e760adf54602e)
![{\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}}}]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/bc72cab8df2afc68e6f8c06531dca45d8ad47d68)
