Question

几个关于数列求和的题目，急用，⒈求Sn=1

Answer 1

let S = 1.a^0+2.a^1+...+n.a^(n-1) (1) aS = 1.a^1+2.a^2+...+n.a^n (2) (2)-(1) (a-1)S =n.a^n -( 1+a+a^2+...+a^(n-1)) =n.a^n -(a^n -1)/(a-1) S =[1/(a-1)^2][ (1-a)n.a^n - (a^n -1) ] an = (3n-1).a^(n-1) = 3(n.a^(n-1)) - a^(n-1) Sn = a1+a2+...+an =3S -(a^n -1)/丛亩(a-1) =[3/(a-1)^2][ (1-a)n.a^n - (a^n -1) ] -(a^n -1)/辩郑盯携和(a-1) =[1/(a-1)^2].[ 2(1-a) n.a^n - 3(a^n -1) ] =[1/(a-1)^2].( 3+ (2(1-a) n- 3) .a^n )