Question

設(shè)數(shù)列{a_n}的前n項(xiàng)和為S_n且S₁=3，若對(duì)任意的n∈N^*都有S_n=2a_n-3n.

(1)求數(shù)列{a_n}的首項(xiàng)及遞推關(guān)系式a_n+1=f(a_n);

(2)求{a_n}的通項(xiàng)公式;

(3)求數(shù)列{a_n}的前n項(xiàng)和S_n.

Answer 1

解：(1)令n=1,S₁=2a₁-3,∴a₁=3.

又S _n+1=2a _n+1-3(n+1),S_n=2a_n-3n，

兩式相減得a _n+1=2a _n+1-2a_n-3,

∴a _n+1=2a_n+3.

(2)a _n+1=2a_n+3a _n+1+3=2(a_n+3),即=2.

∴{a_n+3}是公比為2的等比數(shù)列，其首項(xiàng)為a₁+3=6.

∴a_n+3=(a₁+3)·2^n-1=6·2^n-1.

∴a_n=6·2^n-1-3.

(3)S_n=a₁+a₂+…+a_n

=(6·2⁰-3)+(6·2-3)+(6·2²-3)+…+(6·2^n-1-3)

=(6·2⁰+6·2¹+6·2²+…+6·2^n-1)-(3+3+…+3)

=6(2⁰+2¹+2²+…+2^n-1)-3n

=6·-3n=6·2ⁿ-3n-6.