Question

-1的n次方乘以n的平方,数列求和

Answer 1

∵a[n]=(-1)^n*n^2

∴S[n]=-1+4-9+16-25+36-...-(2k-1)^2+(2k)^2-...+(-1)^n*n^2 (k为正整数)

=3+7+11+...+(4k-1)+...+(-1)^n*n^2

∵当n=2k-1时,k-1=(n-1)/2

∴S[2k-1]=(k-1)[3+4(k-1)-1]/2-(2k-1)^2=(k-1)(2k-1)-(2k-1)^2

即：S[n]=n(n-1)/2-n^2=n^2/2-n/2-n^2=-n(n+1)/2

∵当n=2k时,k=n/2

∴S[2k]=k[3+4k-1]/2=k(2k+1)

即：S[n]=n(n+1)/2

综上所述：S[n]=[(-1)^n]n(n+1)/2