已知数列an是公差大于零的等差数列,数列{bn}为等比数列,且a1=1,b1=2,b2-a2=1,a3+b3=13求数列{an}和{bn}的通项公式设cn=anbn,求数列{cn}的前n项和Tn

Question

已知数列an是公差大于零的等差数列,数列{bn}为等比数列,且a1=1,b1=2,b2-a2=1,a3+b3=13
求数列{an}和{bn}的通项公式
设cn=anbn,求数列{cn}的前n项和Tn

Answer 1

a(n) = 1 + (n-1)d,d>0.
b(n) = 2q^(n-1).
1 = b(2) - a(2) = 2q - (1+d),d = 2q-2.由01.
13 = a(3) + b(3) = 1 + 2d + 2q^2 = 1 + 2(2q-2) + 2q^2,
0 = 2q^2 + 4q - 16 = 2(q^2 + 2q - 8) = 2(q+4)(q-2),q = 2.
d = 2(q-1) = 2.
a(n) = 1 + 2(n-1) = 2n-1.
b(n) = 2^n.
c(n) = a(n)b(n) = (2n-1)2^n.
t(n) = c(1) + c(2) + ...+ c(n-1) + c(n)
= (2*1-1)2 + (2*2-1)2^2 + ...+ [2(n-1)-1]2^(n-1) + (2n-1)2^n,
2t(n) = (2*1-1)2^2 + ...+ [2(n-1)-1]2^n + (2n-1)2^(n+1),
t(n) = 2t(n) - t(n) = -(2*1-1)2 -2*2^2 - ...- 2*2^n + (2n-1)2^(n+1)
= (2n-1)2^(n+1) + 4 - 2 - 2*2 - 2*2^2 - ...- 2*2^n
= (2n-1)2^(n+1) + 4 - 2[1 + 2 + 2^2 + ...+ 2^n]
= (2n-1)2^(n+1) + 4 - 2[2^(n+1) - 1]/(2-1)
= (2n-1)2^(n+1) + 4 - 2*2^(n+1) + 2
= (2n-3)2^(n+1) + 6