Question

設數(shù)列{a_n}的前n項和為S_n,并且a₁=1,S_n+1=4a_n+5(n∈N^*).

(1)求a₃的值;

(2)設b_n=a_n+1-2a_n,求證:數(shù)列{b_n}是等比數(shù)列;

(3)求數(shù)列{a_n}的通項公式.

Answer 1

解析: (1) 由S₂=4a₁+5=9, 又S₂=a₁+a₂,故a₂＝8.再由S₃=9+a₃=4a₂+5=37,得a₃=28.

(2) a_n+1=S_n+1-S_n,a_n+1=4a_n+5-(4a_n-1+5)=4a_n=4a_n-1. 故a_n+1-2a_n=2a_n-4a-1,于是b_n=2b_n-1(n2).又b₁=a₂-2a₁=6,{b_n|是首項為6,公比為2的等比數(shù)列.

(3)由（2）知b_n=6·2^n-1,設cn=,于是c_n+1-c_n==(a_n+1-2a_n)

=·b_n

=·6·2^n-1=.

因此｛c_n|是公差為的等差數(shù)列.

故c₁=+(n-1)·

=n-1.a_n=2ⁿ·c_n=(-1)·2ⁿ=(3n-2)·2^n-1(nN^*).