Question

求数列{[(-1)^(n+1)]*n^2}的前n项和Sn

求数列{ [(-1)^(n+1)] * n^2 }的前n项和Sn

Answer 1

（1）当n为偶数时，令n=2k，则k=n/2
Sn=1²-2²+3²-4²+……+(2k-1)²-(2k)²
=(1-2)(1+2)+(3-4)(3+4)+……+(2k-1-2k)(2k-1+2k)
=-1-2-3-4-……-(2k-1)-2k
=-(2k+1)*2k/2
=-k(2k+1)
=-n(n+1)/2
（2）当n为奇数时，令n=2k-1，则k=(n+1)/2
Sn=1²-2²+3²-4²+……+(2k-3)²-(2k-2)²+(2k-1)²
=(1-2)(1+2)+(3-4)(3+4)+……+(2k-3-2k+2)(2k-3+2k-2)+(2k-1)²
=-1-2-3-4-……-(2k-3)-(2k-2)+(2k-1)²
=-(2k-1)*(2k-2)/2+(2k-1)²
=k(2k-1)
=n(n+1)/2
综上所述，
Sn=(-1)^(n+1)*n(n+1)/2