Question

等差数列×等差数列的前n项和怎么算

Answer 1

用裂项相消法.等差数列×等差数列,通项是关于n的二次式,它可以表示成两个三次式的差.
如：an=n,bn=n+1,
则 an·bn=n²+n
由于 (n+1)³-n³=3n²+3n+1,
即 (n+1)³-n³=3an·bn+1
将n=1,2,3,...代入,得
2³-1³=3a1·b1+1
3³-2³=3a2·b2+1
4³-3³=3a3·b3+1
.
(n+1)³-n³=3an·bn+1
这n个式子相加,得
(n+1)³ -1=3(a1·b1+a2·b2+...+an·bn) +n
即 3Sn=(n+1)³-(n+1)
Sn=n(n+1)(n+2)/3

Answer 2

a(n)=a1+(n-1)d
Sn=na1+n*(n-1)d/2

Answer 3

Sn=nα1+n(n-1)*d/2，a1为等差数列的首项，d为公差