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！

Answer 1

1，a1+a2+a3=3a1+3d=12
∴d=2,an=2n
2,
Sn=2x^1+4x^2+……+2nx^n ①
x*Sn=2x^2+4x^3+……+2nx^(n+1) ②
②-①得(x-1)*Sn=2nx^(n+1)-2(x^1+x^2+……+x^n)=2nx^(n+1)-[x(x^n)-1]/(x-1)

Answer 2

1、a2=a1+d，a3=a1+2d，所以3a1+3d=a1+a2+a3=12，3d=12-3a1=12-6=6
d=2，所以an=a1+(n-1)d=2n
2、bn=2n(x^n)
xSn-Sn=2x+4x²+6(x^3)+....+2n(x^n))(x-1)
所以：Sn=2n(x^(n+1))/(x-1)-(x^(n+1)-x)/(x-1)²

Answer 3

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 4

1,设公差为d,则a1+a2+a3=a2-d+a2+a2+d=3a2=12,a2=4,d=a2-a1=4-2=2
即通项公式为an=2*n
2,bn=2nx^n,b1=2x,b2=4x^2,b3=6x^3……Sn=2x(nx^n-x^(n-1)-x^(n-2)-……-1)/(x-1)

Answer 5

(1)
3a2=12;
a2=4;
a1=2;
an=2n;
(2)
bn=2nX^n
sn=2(X^1+2X^2+3X^3+···nX^n)
=2(X^1+2X^2+3X^3+···nX^n)(1-X) / (1-X) [即分子分母同乘1-X]
=[X-(1+n)X^(n+1)+nX^(n+2)] / (1-X)^2

Answer 6

(1)a1+a1+d+a1+2d=12
a1=1代入得d=3,
an=a1+(n-1)d=1+3(n-1)
=3n-2（n≥1,且n为正整数）
第二题是等比等差数列，好像用错位相减，具体怎么做我不太记得了