(2013•遼寧)下列關(guān)于公差d>0的等差數(shù)列{an}的四個(gè)命題:
p1:數(shù)列{an}是遞增數(shù)列;
p2:數(shù)列{nan}是遞增數(shù)列;
p3:數(shù)列{
an
n
}
是遞增數(shù)列;
p4:數(shù)列{an+3nd}是遞增數(shù)列;
其中真命題是(  )
分析:對(duì)于各個(gè)選項(xiàng)中的數(shù)列,計(jì)算第n+1項(xiàng)與第n項(xiàng)的差,看此差的符號(hào),再根據(jù)遞增數(shù)列的定義得出結(jié)論.
解答:解:∵對(duì)于公差d>0的等差數(shù)列{an},an+1-an=d>0,∴命題p1:數(shù)列{an}是遞增數(shù)列成立,是真命題.
對(duì)于數(shù)列數(shù)列{nan},第n+1項(xiàng)與第n項(xiàng)的差等于 (n+1)an+1-nan=nd+an+1,不一定是正實(shí)數(shù),
故p2不正確,是假命題.
對(duì)于數(shù)列{
an
n
}
,第n+1項(xiàng)與第n項(xiàng)的差等于 
an+1
n+1
-
an
n
=
nan+1-(n+1)an
n(n+1)
=
nd-an
n(n+1)
,不一定是正實(shí)數(shù),
故p3不正確,是假命題.
對(duì)于數(shù)列數(shù)列{an+3nd},第n+1項(xiàng)與第n項(xiàng)的差等于 an+1+3(n+1)d-an-3nd=4d>0,
故命題p4:數(shù)列{an+3nd}是遞增數(shù)列成立,是真命題.
故選D.
點(diǎn)評(píng):本題主要考查等差數(shù)列的定義,增數(shù)列的含義,命題的真假的判斷,屬于中檔題.
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