Question

设数列an的前n项和为Sn=2n∧2,bn为等比数列,且a1=b1,b2(a2-a3)=b1(1)求数列an和bn的通项

Answer 1

a(1)=2,

a(n+1)=s(n+1)-s(n)=2(2n+1)=4n+2,

a(n)=4(n-1)+2.

b(n)=2q^(n-1),

2=b(1)=b(2)[a(2)-a(3)]=2q[-4],

q=-1/4.

b(n)=2(-1/4)^(n-1).

c(n)=a(n)/b(n)=[4(n-1)+2]/[2(-1/4)^(n-1)]=[2(n-1)+1](-4)^(n-1),

t(n)=[2(1-1)+1] + [2(2-1)+1](-4) + [2(3-1)+1](-4)^2 + ...+ [2(n-1-1)+1](-4)^(n-2) + [2(n-1)+1](-4)^(n-1),

-4t(n)=[2(1-1)+1](-4) + [2(2-1)+1](-4)^2 + ...+[2(n-1-1)+1](-4)^(n-1) + [2(n-1)+1](-4)^n

5t(n)=t(n)-[-4t(n)]=[2(1-1)+1] + [2](-4)+[2](-4)^2+...+[2](-4)^(n-1) - [2(n-1)+1](-4)^n

=-1+2[1+(-4)+(-4)^2+...+(-4)^(n-1)] -(2n-1)(-4)^n

=-1-(2n-1)(-4)^n + 2[1-(-4)^n]/[1-(-4)]

=-3/5 -(2n-1)(-4)^n - (2/5)(-4)^n

t(n)=-3/25 - [2n-3/5](-4)^n/5