Question

各项均为正数的等比数列｛an｝,满足a1=1,a(n+1)²-an²=2(n∈N+)； (1)求｛

Answer 1

(1)

[a(n+1)]^2-(an)^2 =2

{ (an)^2} 是等差数列,d=2

(an)^2 - (a1)^2 = 2(n-1)

(an)^2 = 2n-1

an = √(2n-1)

(2)

(an)^2/2^n = (2n-1)(1/2^n)

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

Sn = S - ( 1- 1/2^n)

S = 1.(1/2)^0 + 2.(1/2)^1 +...+ n(1/2)^(n-1) (1)

(1/2)S = 1.(1/2)^1 + 2.(1/2)^2 +...+ n(1/2)^n (2)

(1)-(2)

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

= 2(1- 1/2^n)- n(1/2)^n

S = 4(1- 1/2^n)- 2n(1/2)^n

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

Sn = S - ( 1- 1/2^n)

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

= 3 -(2n+3)(1/2)^n