Question

来个大神！！ 求数列(2n-1)²分之一的前n项和。

求大佬写详细点。谢谢

Answer 1

an
=(2n-1)^2
=4n^2 -4n +1
=4n(n-1) +1
=(4/3)[ (n-1)n(n+1) - (n-2)(n-1)n] + 1
Sn
= a1+a2+...+an
=(4/3)(n-1)n(n+1) +n
=(1/3)n [ 4(n-1)(n+1) + 3 ]
=(1/3)n(4n^2-4n + 3 )
=(1/3)n(4n^2-4n + 3 )

Answer 2

来个大神！！

追问

。。。

Answer 3

But the car didn't decelerate