Question

已知数列{an}是等差数列,且a1=2,a1+a2+a3=12 (1)求数列{an}的通项公式. (2)令bn=anX^n(x不等于1),求数...

已知数列{an}是等差数列,且a1=2,a1+a2+a3=12
(1)求数列{an}的通项公式.
(2)令bn=anX^n(x不等于1),求数列{bn}的前n项和公式sn.
要详解答案,我采纳！急用呀！

Answer 1

a1+a2+a3=3a1+3d=6+3d=12
d=2
an=a1+(n-1)d=2+2(n-1)=2n
bn=anx^n=2nx^n
Sn=2(x+2x^2+3x^3+...+nx^n)
Sn/x=2(1+2x+3x^2+...+nx^(n-1)
Sn=2nx^(n+1)/(1-x)-2x(x^n-1)/(x-1)^2

Answer 2

(1) a1+a2+a3=3a1+3d=12,d=2,an=a1+2(n-1)
(2)bn=[a1+2(n-1)]x^n.Sn=a1*x+(a1+d)x^2+(a1+2d)x^3+……+[a1+d(n-1)]x^n=a1(x+x^2+...+x^n)+(dx^2+d^2x^3+...+d^(n-1)x^n)=a1[x(1-x^n)/(1-x) ]+dx^2[1-(dx)^n]/(1-dx)