Question

已知数列｛an}满足a1，a2-a1,a3-a1,...,an-an-1是首项为1，公比为1/3的等比数列。

求an;若bn=(2n-1)an,求｛bn｝的前n项和Sn

Answer 1

a(n+1)-a(n)=(1/3)^(n-1),
3^na(n+1)=3*3^(n-1)a(n)+3,
3^na(n+1)+3/2 = 3*3^(n-1)a(n)+3+3/2=3[3^(n-1)a(n)+1+1/2]=3[3^(n-1)a(n)+3/2]
{3^(n-1)a(n)+3/2}是首项为a(1)+3/2=5/2, 公比为3的等比数列.
3^(n-1)a(n)+3/2=(5/2)*3^(n-1),
a(n) + (3/2)/3^(n-1) = 5/2,
a(n) = 5/2 - (3/2)(1/3)^(n-1).

b(n)=(2n-1)a(n)=5(2n-1)/2 - 3(2n-1)/2*(1/3)^(n-1)=(5/2)(2n-2+1) - (3/2)(2n-1)(1/3)^(n-1)
=5(n-1)+5/2 - (3/2)(2n-1)(1/3)^(n-1),
设c(n)=(2n-1)(1/3)^(n-1),
t(n)=c(1)+c(2)+...+c(n-1)+c(n)
=1/1 + 3/3 + ... + [2(n-1)-1]/3^(n-2) + (2n-1)/3^(n-1),
3t(n) = 3/1 + 3/1 + ... + [2(n-1)-1]/3^(n-3) + (2n-1)/3^(n-2),
2t(n)=3t(n)-t(n)=3/1 + 2/1 + ... + 2/3^(n-2) - (2n-1)/3^(n-1)
=3 + 2[1+1/3+...+1/3^(n-2)] - (2n-1)/3^(n-1)
=3+2[1-1/3^(n-1)]/(1-1/3) -(2n-1)/3^(n-1)
=3+3[1-1/3^(n-1)] - (2n-1)/3^(n-1)
=6-(2n+2)/3^(n-1),
t(n)=3-(n+1)/3^(n-1)

s(n)=b(1)+b(2)+...+b(n-1)+b(n)
=5n(n-1)/2 + 5n/2 - (3/2)[c(1)+c(2)+...+c(n-1)+c(n)]
=5n(n-1)/2 + 5n/2 -(3/2)t(n)
=5n(n-1)/2 + 5n/2 - (3/2)[3-(n+1)/3^(n-1)]
=5n(n-1)/2 + 5n/2 - 9/2 + (3/2)(n+1)/3^(n-1)