Question

設(shè)數(shù)列{an}滿足a1=2,a2+a4=8,且對任意n∈N*,函數(shù)f(x)=(an-an+1+an+2)x+an+1cos x-an+2sin x滿足

f′=0.

(1)求數(shù)列{an}的通項(xiàng)公式;

(2)若bn=2(an+),求數(shù)列{bn}的前n項(xiàng)和Sn.

Answer 1

解:(1)由題設(shè)可得,

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

對任意n∈N*,f′=an-an+1+an+2-an+1=0,

即an+1-an=an+2-an+1,故{an}為等差數(shù)列.

由a1=2,a2+a4=8,

解得數(shù)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-.