Question

等比数列问题求解十二题

Answer 1

这样

追问

第二步是什么来的

说错了，第三步

Answer 2

供参考。

追问

第二步是什么来的

追答

平方差公式

Answer 3

let
bn
= n^2
=n(n+1) -n
=(1/3)[n(n+1)(n+2) - (n-1)n(n+1)] - (1/2)[n(n+1) -(n-1)n]
Tn =b1+b2+...+bn
=(1/3)n(n+1)(n+2) - (1/2)n(n+1)
=(1/6)n(n+1)(2n+1)

an =(-1)^n . n^2
if n is odd
an =(-1)^n . n^2
=-(b1+b2+....+bn) + 2[b2+b4+....+b(n-1)]
=-(1/6)n(n+1)(2n+1) + 8( 1^2+2^2+...+ [(n-1)/2]^2 )
=-(1/6)n(n+1)(2n+1) + 8(1/6)[(n-1)/2][(n+1)/2](n)
=-(1/6)n(n+1)(2n+1) + (1/3)(n-1)(n+1)n
=-(1/6)n(n+1)( 2n+1 -2n+2)
= -(1/2)n(n+1)
if n is even
an =(-1)^n . n^2
=-(b1+b2+....+bn) + 2[b2+b4+....+bn]
=-(1/6)n(n+1)(2n+1) + 8( 1^2+2^2+...+ (n/2)^2 )
=-(1/6)n(n+1)(2n+1) + 8(1/6)(n/2)[(n+3)/2](n+1)
=-(1/6)n(n+1)(2n+1) + (1/3)n(n+1)(n+3)
=- (1/6)n(n+1)( 2n+1 -2n-6)
= (5/6)n(n+1)