已知數(shù)列{an}對(duì)所有正整數(shù)n滿足an<an+1,且an=2n2+pn,則實(shí)數(shù)p的取值范圍是( )
A.(-∞,-6)
B.(-6,+∞)
C.(-∞,6)
D.(6,+∞)
【答案】分析:已知數(shù)列{an}中,an=n2+λn,且an是遞增數(shù)列,求實(shí)數(shù)λ的取值范圍,根據(jù)所給的數(shù)列的項(xiàng),寫出數(shù)列的第n+1項(xiàng),根據(jù)數(shù)列滿足an<an+1,把所給的兩項(xiàng)做差,得到不等式,根據(jù)恒成立得到結(jié)果.
解答:解:∵an=2n2+pn,
∴an+1=2(n+1)2+p(n+1)
∵數(shù)列{an}對(duì)所有正整數(shù)n滿足an<an+1,
∴2(n+1)2+p(n+1)-2n2-pn>0
即4n+2+p>0
∴p>-4n-2
∵對(duì)于任意正整數(shù)都成立,
∴p>-6
則實(shí)數(shù)p的取值范圍是:(-6,+∞)
故選B.
點(diǎn)評(píng):本題考查數(shù)列的函數(shù)的特性,本題解題的關(guān)鍵是防寫出數(shù)列的一項(xiàng),根據(jù)函數(shù)的思想,得到不等式且解出不等式.