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

(1) a_n=n+1    (2) S_n=n²+3n+1-

解析解:(1)由题设可得,
f′(x)=a_n-a_n+1+a_n+2-a_n+1sin x-a_n+2cos x.
对任意n∈N^*,f′=a_n-a_n+1+a_n+2-a_n+1=0,
即a_n+1-a_n=a_n+2-a_n+1,故{a_n}为等差数列.
由a₁=2,a₂+a₄=8,
解得数列{a_n}的公差d=1,
所以a_n=2+1×(n-1)=n+1.
(2)由b_n=2(a_n+)=2(n+1+)=2n++2知,
S_n=b₁+b₂+…+b_n
=2n+2·+
=n²+3n+1-.