数列{an}的前n项和Tn，且an=(-1)的n次方╳n的平方，求T2n, T2n+1

Question

如题所述

Answer 1

an= (-1)^n . n^2
T2n = a1+a2+...+a2n
= [a1+a3+...+a(2n-1)] + [ a2+a4+...+a2n]
=-[1^2+3^2+...+(2n-1)^2] +[2^2+4^2+...+(2n)^2]
= -[1^2+2^2+...+(2n)^2] + 2[2^2+4^2+...+(2n)^2]
= -[1^2+2^2+...+(2n)^2] + 8(1^2+2^2+...+n^2)
= -(1/6)(2n)(2n+1)(4n+1) + (8/6)n(n+1)(2n+1)
= -(1/3)n(2n+1)(4n+1) + (4/3)n(n+1)(2n+1)
=(1/3)n[ 4(n+1)(2n+1)- (2n+1)(4n+1) ]
=(1/3)n [ 4(2n^2+3n+1) -(8n^2+6n+1) ]
=n(2n+1)
T(2n+1)= a1+a2+...+a(2n+1)
=[a1+a3+...+a(2n+1)] + [ a2+a4+...+a2n]
=-[1^2+3^2+...+(2n+1)^2] +[2^2+4^2+...+(2n)^2]
= -[1^2+2^2+...+(2n+1)^2] + 2[2^2+4^2+...+(2n)^2]
= -[1^2+2^2+...+(2n+1)^2] + 8[1^2+2^2+...+n^2]
=-(1/6)(2n+1)(2n+2)(4n+3) + (8/6)n(n+1)(2n+1)
=-(1/3)(2n+1)(n+1)(4n+3) + (4/3)n(n+1)(2n+1)
= (1/3)(n+1)(2n+1)[ 4n- (4n+3) ]
= -(n+1)(2n+1)