Question

已知數(shù)列{a_n}的首項(xiàng)a₁=5,前n項(xiàng)和為S_n,且S_n+1=S_n+n+5(n∈N^*).

(1)證明數(shù)列{a_n+1}是等比數(shù)列;

(2)令f(x)=a₁x+a₂x²+…+a_nxⁿ,求函數(shù)f(x)在點(diǎn)x=1處的導(dǎo)數(shù)f′(1)并比較2f′(1)與23n²-13n的大小.

Answer 1

解:(1)由已知S_n+1=S_n+n+5(n∈N^*)可得n≥2,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),S₂=2S₁+1+5,則a₂+a₁=2a₁+6.

又a₁=5,所以a₂=11.從而a₂+1=2(a₁+1).

故總有a_n+1+1=2(a_n+1),n∈N^*.

又a₁=5,a₁+1≠0,從而=2,即數(shù)列{a_n+1}是等比數(shù)列.

(2)由(1)知a_n=3×2ⁿ-1._?

因?yàn)閒(x)=a₁x+a₂x²+…+a_nxⁿ,所以f′(x)=a₁+2a₂x+…+na_nx^n-1._?

從而f′(1)=a₁+2a₂+…+na_n=(3×2-1)+2(3×2²-1)+…+n(3×2ⁿ-1)=3(2+2×2²+…+n×2ⁿ)-(1+2+…+n)=3(n-1)·2ⁿ⁺¹-+6._?

由上得2f′(1)-(23n²-13n)=12(n-1)·2ⁿ-12(2n²-n-1)=12(n-1)·2ⁿ-12(n-1)(2n+1)=12(n-1)［2ⁿ-(2n+1)］.①_?

當(dāng)n=1時(shí),①式=0,所以2f′(1)=23n²-13n;

當(dāng)n=2時(shí),①式=-12＜0,所以2f′(1)＜23n²-13n;

當(dāng)n≥3時(shí),n-1＞0,又2ⁿ=(1+1)ⁿ=C⁰_n+C¹_n+…+C^n-1_n+Cⁿ_n≥2n+2＞2n+1,_?

所以(n-1)［2ⁿ-(2n+1)］＞0,即①＞0.從而2f′(1)＞23n²-13n._?

(或用數(shù)學(xué)歸納法，略).