Question

设数列AN满足A1=2,A(N+1)-AN=3X2^(2N-1)？

(1)求数列AN的通项公式
2，令BN=N AN ,求BN前N项和SN

Answer 1

a(n+1)-an=3*2^(2n-1)
an-a(n-1)=3*2^(2n-3)
...
a3-a2=3*2^3
a2-a1=3*2^1
相加
an-a1=3[2^1+2^3+2^5+2^7+...+2^(2n-3)]
=3*2*[2^(2n-2)-1]/(2^2-1)
=2^(2n-1)-2

an=a1+2^(2n-1)-2=2^(2n-1)

an=2^(2n-1)

bn=n*2^(2n-1)

Answer 2

(1)
a2-a1=3*2^(2*1-1)=(3/2)*4
a3-a2=3*2^(2*2-1)=(3/2)*4^2
a4-a3=...=(3/2)*4^3
...
an-a(n-1)=(3/2)*4^(n-1)

相加有

a(n)-a1=(3/2)*(4+4^2+4^3+...+4^(n-1))
=(3/2)*(4/3)*(4^(n-1)-1)
=2*(4^(n-1)-1)

a(n)=2+2*(4^(n-1)-1)=2^(2n-1)

(2)
Bn=n*an=n*2^(2n-1)

Sn=1*2^1+2*2^3+3*2^5+4*2^7+...+(n-1)*2^((2n-3)+n*2^(2n-1)

4Sn=2^2*Sn=1*2^3+2*2^5+3*2^7+....+(n-1)*2^(2n-1)+n*2^(2n+1)

-3Sn=[2^1+2^3+2^5+2^7+...+2^(2n-1)]-2*2^(2n+1)
=(2/3)*2^2n-1)-2*2^(2n+1)

Sn=...

Answer 3

an=2^(2n-1)
bn=n*2^(2n-1)
s(n)=1*2^1+2*2^3+……+n*2^(2n-1)两边同时乘以4
4s(n)=1*2^3+2*2^5+……+n*2^(2n+1)两式相减
3s(n)=-2^1-2^3-……-2^(2n-1)+n*2^(2n+1)
s(n)=[(3n-1)*2^(2n+1)+2]/9