发布网友 发布时间:2024-10-24 13:22
共1个回答
热心网友 时间:2024-10-27 04:14
(1)原式=(a+a^2+……+a^n)-(1+2+……+n)
=a[1+a+……+a^(n-1)]-n(n+1)/2
=a(1-a^n)/(1-a)-n(n+1)/2
(2)原式=(2+4+……+2n)-3(1/5+1/5^2+……+1/5^n)
=2(1+2+……+n)-(3/5)[1+1/5+……+1/5^(n-1)]
=n(n+1)-(3/5)(1-1/5^n)/(1-1/5)
=n(n+1)-(3/4)(1-1/5^n)
(3)设S=1+2x+3x^2+……+nx^(n-1)①
xS=x+2x^2+……+(n-1)x^(n-1)+nx^n②
①-②: S-xS=[1+x+x^2+……+x^(n-1)]-nx^n
∴(1-x)S=(1-x^n)/(1-x)-nx^n
=(1-x^n-nx^n+nx^(n+1)]=[1-(n+1)x^n+nx^(n+1)]
∴S=[1-(n+1)x^n+nx^(n+1)]/(1-x)