数列{(-1)^(n-1)*n^2}的前n项之和为

Question

数列{(-1)^(n-1)*n^2}的前n项之和为

Answer 1

通项an=(-1)^(n-1)*n^2
1)若n=2k,k=0,1,2,3,...
则an=(-1)^(2k-1)*(2k)^2
=-(2k)^2
第n-1项为：(-1)^(2k-2)*(2k-1)^2
=(2k-1)^2
该两项之和为：-(2k)^2+(2k-1)^2
=-[(2k-1)+2k]
=-[(n-1)+n]
故而前n项和为Sn=-[1+2+3+...+(n-1)+n]
=-(1+n)*n/2
=-k(2k+1)
2)若n=2k+1,k=0,1,2,3,...
则前n项和为：-k(2k+1)+(-1)^(n-1)*n^2
=-k(2k+1)+(-1)^(2k+1-1)*(2k+1)^2
=-k(2k+1)+(2k+1)^2
=(k+1)(2k+1)
=(1+n)*n/2
综上所述,若n为偶数则前n项和为-(1+n)*n/2
若n为奇数则前n项和为(1+n)*n/2