Question

已知等差数列an的通项公式为an=1+2n，令bn=an的平方-1，求bn的前n项和

是bn=an的平方-1分之1 ，题上的错了

Answer 1

an=1+2n
bn=（2n+1)^2-1=4n^2+4n
设bn的项和为sn，4n^2的n项和为An，4n的n项和为Bn
sn=An+Bn
=2n（n+1）（2n+1）/3+2n(2n+1)

追问

中间步骤那

Answer 2

答：
等差数列An=1+2n
Bn=(An)^2-1
=(An -1)(An +1)
=2n(2n+2)
=4n(n+1)
=4n^2+4n
Sn=4*[(1^2+2^2+3^2+...n^2) +(1+2+3+...+n)]
=4*[ n(n+1)(2n+1)/6+n(n+1)/2]
=4n(n+1)*(2n+1+3)/6
=4n(n+1)(n+2)/3

Answer 3

an=2n+1
(an)^2=4n^2+4n+1
bn=1/[(an)^2-1]
= 1/[4n(n+1)]
=(1/4)[1/n-1/(n+1) ]
Sn=b1+b2+...+bn
=(1/4)[ 1- 1/(n+1)]
= n/[4(n+1)]

Answer 4