Question

设数列{an}满足a1=2,a(n+1)-an=3乘以2的（2n-1}次方 1.求数列的通项公式； 2.令bn=n乘以an,求数列前n项和

Answer 1

a1=2
a2-a1=3*2^(2-1)=6
令cn=a(n+1)-an=3*2^(2n-1),则c1=a2-a1=6,cn/c(n-1)=4
cn是首项是6公比是4的等比数列
设cn的前n-1项和为s(n-1)
则s(n-1)=an-a(n-1)+......+a2-a1=an-a1=6*[1-4^(n-1)]/(1-4)=2*[4^(n-1)-1]
an=2*[4^(n-1)-1]+2=2*[4^(n-1)]=2^(2n-1)
即通项是 an=2^(2n-1)
Tn=b1+.....+bn=2+2*2^3+3*2^5+........+n*2^(2n-1)
2^2*Tn=4Tn= 2^3+2*2^5+...+(n-1)*2^(2n-1)+n*2^(2n+1)
-3Tn=Tn-4Tn=2+2^3+2^5+.....+2^(2n-1)-n*2^(2n+1)=2*(1-4^n)/(1-4)-n*2^(2n+1)
=[(1-3n)*2^(2n+1)-2]/3
Tn=[(3n-1)*2^(2n+1)-2]/9

Answer 2

a1=2
a2-a1=3*2^(2-1)=6
令cn=a(n 1)-an=3*2^(2n-1),则c1=a2-a1=6,cn/c(n-1)=4
cn是首项是6公比是4的等比数列
设cn的前n-1项和为s(n-1)
则s(n-1)=an-a(n-1) ...... a2-a1=an-a1=6*[1-4^(n-1)]/(1-4)=2*[4^(n-1)-1]
an=2*[4^(n-1)-1] 2=2*[4^(n-1)]=2^(2n-1)
即通项是 an=2^(2n-1)
Tn=b1 ..... bn=2 2*2^3 3*2^5 ........ n*2^(2n-1)
2^2*Tn=4Tn= 2^3 2*2^5 ... (n-1)*2^(2n-1) n*2^(2n 1)
-3Tn=Tn-4Tn=2 2^3 2^5 ..... 2^(2n-1)-n*2^(2n 1)=2*(1-4^n)/(1-4)-n*2^(2n 1)=[(1-3n)*2^(2n 1)-2]/3
Tn=[(3n-1)*2^(2n 1)-2]/9