Question

已知數(shù)列{a_n}的首項(xiàng)a₁=2,點(diǎn)(a_n,a_n+1)在函數(shù)f(x)=x²+2x的圖象上,其中n∈N₊.

(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

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

又∵a₁=2,∴a_n+1＞0,兩邊取對(duì)數(shù)得lg(a_n+1+1)=2lg(a_n+1)(n∈N₊),即=2.

∴數(shù)列{lg(1+a_n)}是公比為2的等比數(shù)列.

(2)由(1)知lg(1+a_n)=2n-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)=··…=+…+2^n-1=3^(2n-1).

由(*)式得a_n=-1.

(3)∵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().

又∵a_n=-1,a₁=2,a_n+1=-1,∴S_n=1.

又由(2)知T_n=,∴S_n+=1+=1.