Question

求x^2+x^4+x^6+2y+4y+6y+.2n项的和(等差和等比级数)

Answer 1

分别求和

x^2+x^4+x^6+……+x^(2n)

q=x^2,a1=x^2

所以x^2+x^4+x^6+……+x^(2n)

=x^2*[(x^2)^n-1]/(x^2-1)

=x^2*[x^(2n)-1]/(x^2-1)

2y+4y+6y+……+2ny

一共n项

a1=2y,an=2ny

所以2y+4y+6y+……+2ny

=(2y+2ny)*n/2

=yn(n+1)

所以x^2+x^4+x^6+……+x^(2n)+2y+4y+6y+……+2ny

=x^2*[x^(2n)-1]/(x^2-1)+yn(n+1)