Question

設(shè)函數(shù)f(x)=+sinx的所有正的極小值點(diǎn)從小到大排成的數(shù)列為{xn}.

(1)求數(shù)列{xn}的通項(xiàng)公式.

(2)設(shè){xn}的前n項(xiàng)和為Sn,求sinSn.

 

Answer 1

(1) xn=2nπ-(n∈N*) (2) sinSn=

【解析】【思路點(diǎn)撥】(1)根據(jù)導(dǎo)數(shù),xn的左側(cè)導(dǎo)函數(shù)小于0,xn的右側(cè)導(dǎo)函數(shù)大于0,求出極小值點(diǎn).(2)由(1)求出{xn}的前n項(xiàng)和為Sn,再代入sinSn求解.

【解析】
(1)f(x)=+sinx,令f'(x)=+cosx=0,得x=2kπ±(k∈Z),

f'(x)>0⇒2kπ-<x<2kπ+(k∈Z),

f'(x)<0⇒2kπ+<x<2kπ+(k∈Z),

當(dāng)x=2kπ-(k∈Z)時(shí),f(x)取極小值,

xn=2nπ-(n∈N*).

(2)由(1)得:xn=2nπ-,

Sn=x1+x2+x3+…+xn

=2π(1+2+3+…+n)-=n(n+1)π-.

當(dāng)n=3k(k∈N*)時(shí),sinSn=sin(-2kπ)=0,

當(dāng)n=3k-1(k∈N*)時(shí),sinSn=sin=,

當(dāng)n=3k-2(k∈N*)時(shí),sinSn=sin=-.

所以sinSn=