Question

已知a₁=2,點(diǎn)(a_n,a_n+1)在函數(shù)f(x)=x²+2x的圖象上,其中n=1,2,3,….

(1)證明數(shù)列{lg(1+a_n)}是等比數(shù)列;

(2)設(shè)T_n=(1+a₁)(1+a₂)…(1+a_n),求T_n及數(shù)列{a_n}的通項(xiàng);

(3)記b_n=+,求數(shù)列{b_n}的前n項(xiàng)和S_n,并證明S_n+=1.

Answer 1

(1)證明:由已知a_n+1=a_n²+2a_n,∴a_n+1+1=(a_n+1)².

∵a₁=2,∴a_n+1＞1,兩邊取對(duì)數(shù)得lg(1+a_n+1)=2lg(1+a_n),即=2.∴{lg(1+a_n)}是公比為2的等比數(shù)列.

(2)解：由(1)知lg(1+a_n)=2^n-1·lg(1+a₁)=2^n-1·lg3=lg3^2n-1.∴1+a_n=3^2n-1.                     (*)

∴T_n=(1+a₁)(1+a₂)…(1+a_n)=·…·=,由（*）式得a_n=3^2n-1-1.

（3）解：∵a_n+1=a_n²+2a_n,∴a_n+1=a_n(a_n+2).∴.

∴.

又b_n=+,

∴b_n=2().

∴S_n=b₁+b₂+…+b_n=2（）=2().

∵a_n=3^2n-1-1,a₁=2,a_n+1=3^2n-1,∴S_n=1-.又T_n=3^2n-1,∴S_n+=1.