已知f (x)=mx(m為常數(shù),m>0且m≠1).設(shè)f (a1),f (a2),…,f (an),…(n∈N)是首項(xiàng)為m2,公比為m的等比數(shù)列.
(1)求證:數(shù)列{an}是等差數(shù)列;
(2)若bn=an f (an),且數(shù)列{bn}的前n項(xiàng)和為Sn,當(dāng)m=3時(shí),求Sn;
(3)若cn=f(an)lgf (an),問(wèn)是否存在m,使得數(shù)列{cn}中每一項(xiàng)恒不小于它后面的項(xiàng)?若存在,求出m的取值范圍;若不存在,請(qǐng)說(shuō)明理由.
解:(1)由題意f (a
n)=m
2•m
n-1,即m
an=m
n+1.
∴a
n=n+1,∴a
n+1-a
n=1,∴數(shù)列{a
n}是以2為首項(xiàng),1為公差的等差數(shù)列.
(2)由題意b
n=a
nf (a
n)=(n+1)•m
n+1,
當(dāng)m=3時(shí),b
n=(n+1)•3
n+1,∴S
n=2•3
2+3•3
3+4•3
4+…+(n+1)•3
n+1…①,
①式兩端同乘以3得,3S
n=2•3
3+3•3
4+4•3
5+…+(n+1)•3
n+2…②
②-①并整理得,
2S
n=-2•3
2-3
3-3
4-3
5-…-3
n+1+(n+1)•3
n+2=-3
2-(3
2+3
3+3
4+3
5+…+3
n+1)+(n+1)•3
n+2
=-32-
+(n+1)•3
n+2=-9+
(1-3
n)+(n+1)•3
n+2=(n+
)3
n+2-
.
∴S
n=
(2n+1)3
n+2-
.
(3)由題意c
n=f (a
n)•lg f (a
n)=m
n+1•lgm
n+1=(n+1)•m
n+1•lgm,
要使c
n≥c
n+1對(duì)一切n∈N*成立,即(n+1)•m
n+1•lgm≥(n+2)•m
n+2•lgm,對(duì)一切n∈N
*成立,
當(dāng)m>1時(shí),lgm>0,所以n+1≥m(n+2),即m≤
對(duì)一切n∈N
*成立,
因?yàn)?img class='latex' src='http://thumb.zyjl.cn/pic5/latex/20131.png' />=1-
的最小值為
,所以m≤
,與m>1不符合,即此種情況不存在.
②當(dāng)0<m<1時(shí),lgm<0,所以n+1≤m(n+2),即m≥
對(duì)一切n∈N
*成立,所以
≤m<1.
綜上,當(dāng)
≤m<1時(shí),數(shù)列{c
n}中每一項(xiàng)恒不小于它后面的項(xiàng).
分析:(1)利用f (x)=m
x(m為常數(shù),m>0且m≠1).代入a
n,求出a
n的表達(dá)式,利用等差數(shù)列的定義,證明數(shù)列{a
n}是等差數(shù)列;
(2)通過(guò)b
n=a
n f (a
n),且數(shù)列{b
n}的前n項(xiàng)和為S
n,當(dāng)m=3時(shí),求出S
n的表達(dá)式,利用錯(cuò)位相減法求出S
n;
(3)利用c
n=f(a
n)lgf (a
n),要使c
n≥c
n+1對(duì)一切n∈N
*成立,推出m,n的關(guān)系式,通過(guò)m>1,0<m<1結(jié)合一切n∈N
*,數(shù)列{c
n}中每一項(xiàng)恒不小于它后面的項(xiàng),推出m的取值范圍;
點(diǎn)評(píng):本題考查數(shù)列的定義的應(yīng)用,錯(cuò)位相減法,數(shù)列與函數(shù)相結(jié)合,恒成立問(wèn)題的綜合應(yīng)用,考查分析問(wèn)題解決問(wèn)題,轉(zhuǎn)化思想的應(yīng)用,知識(shí)面廣,運(yùn)算量大.