已知數(shù)列{an}的首項(xiàng)a1=5,前n項(xiàng)和為Sn,
且Sn+1=2Sn+n+5(n∈N*).
(Ⅰ)證明數(shù)列{an+1}是等比數(shù)列;
(Ⅱ)令f(x)=a1x+a2x2+…+anxn,求函數(shù)f(x)在點(diǎn)x=1處的導(dǎo)數(shù)f'(1).
分析:(Ⅰ)先根據(jù)S
n+1=2S
n+n+5可得到S
n=2S
n-1+n+4,然后兩式相減可得到S
n+1-S
n=2(S
n-S
n-1)+1,即a
n+1=2a
n+1然后兩邊同時(shí)加1即可得到a
n+1+1=2(a
n+1),即
=2.從而得證.
(Ⅱ)先根據(jù)(Ⅰ)求出a
n的通項(xiàng)公式,再對(duì)函數(shù)f(x)進(jìn)行求導(dǎo),得到f'(x)的表達(dá)式,然后將a
n的表達(dá)式代入進(jìn)行分組求和即可.
解答:解:(Ⅰ)由已知S
n+1=2S
n+n+5,∴n≥2時(shí),S
n=2S
n-1+n+4,
兩式相減,得S
n+1-S
n=2(S
n-S
n-1)+1,
即a
n+1=2a
n+1,從而a
n+1+1=2(a
n+1).
當(dāng)n=1時(shí),S
2=2S
1+1+5,∴a
1+a
2=2a
1+6又a
1=5,∴a
2=11,
從而a
2+1=2(a
1+1).故總有a
n+1+1=2(a
n+1),n∈N*.
又∵a
1=5,,∴a
n+1≠0,從而
=2.
即{a
n+1}是以a
1+1=6為首項(xiàng),2為公比的等比數(shù)列.
(Ⅱ)由(Ⅰ)知a
n=3×2
n-1.
∵f(x)=a
1x+a
2x
2+…+a
nx
n∴f'(x)=a
1+2a
2x+…+na
nx
n-1.
從而f'(1)=a
1+2a
2+…+na
n=(3×2-1)+2(3×2
2-1)+…+n(3×2
n-1)
=3(2+2×2
2+…+n×2
n)-(1+2+…+n)
=
3[n×2n+1-(2+…+2n)]-=
3[n×2n+1-2n+1+2]-=
3(n-1)•2n+1-+6.
點(diǎn)評(píng):本題主要考查等比數(shù)列的證明、求導(dǎo)運(yùn)算和數(shù)列的分組求和.考查基礎(chǔ)知識(shí)的綜合運(yùn)用和計(jì)算能力.