Question

设数列{a_n}满足a₁=2,a₂+a₄=8,且对任意n∈N^*,函数f(x)=(a_n-a_n+1+a_n+2)x+a_n+1cos x-a_n+2sin x满足f′=0.

(1)求数列{a_n}的通项公式;

(2)若b_n=2(a_n+),求数列{b_n}的前n项和S_n.

Answer

(1) an=n+1 (2) Sn=n2+3n+1-

【解析】

解:(1)由题设可得,

f′(x)=an-an+1+an+2-an+1sin x-an+2cos x.

对任意n∈N*,f′=an-an+1+an+2-an+1=0,

即an+1-an=an+2-an+1,故{an}为等差数列.

由a1=2,a2+a4=8,

解得数列{an}的公差d=1,

所以an=2+1×(n-1)=n+1.

(2)由bn=2(an+)=2(n+1+)=2n++2知,

Sn=b1+b2+…+bn

=2n+2·+

=n2+3n+1-.