Question

已知{an}是一个公差大于0的等差数列，且满足a3a6＝55,a2+a7＝16 (1)求数列{an}

已知{an}是一个公差大于0的等差数列，且满足a3a6＝55,a2+a7＝16
(1)求数列{an}的通项公式
(2)若数列{an}和数列bn满足等式：an＝b1/2+b2/2^2+b3/2^3……+bn/2^n(n为正数)，求数列{bn}的前n项和Sn

Answer 1

(1)
an=a1+(n-1)d
a3a6=55
(a1+2d)(a1+5d)=55 (1)
a2+a7=16
2a1+7d=16
a1=(16-7d)/2 (2)
sub (2) into (1)
((16-7d)/2+2d)((16-7d)/2+5d)=55
(16-3d)(16+3d)=220
256-9d^2=220
9d^2-36=0
d^2=4
d=2 or -2(rejected)
a1=1
an=1+(n-1)2= 2n-1
(2)
an= b1/2+b2/2^2+b3/2^3……+bn/2^n
an- a(n-1) = (b1/2+b2/2^2+b3/2^3……+bn/2^n)- (b1/2+b2/2^2+b3/2^3……+b(n-1)/2^(n-1) )
d = bn/2^n
bn = 2(2^n) = 2^(n+1)
Sn = b1+b2+..+bn
= 2^2+2^3+...2^(n+1)
= 2^2(2^n-1)
=4(2^n-1)

Answer 2

额