函數(shù)f(x)的定義域為R,數(shù)列{an}滿足an=f(an-1)(n∈N*且n≥2). (Ⅰ)若數(shù)列{an}是等差數(shù)列,a1≠a2,且f(an)-f(an-1)=k(an-an-1)(k為非零常數(shù),n∈N*且n≥2),求k的值; (Ⅱ)若f(
函數(shù)f(x)的定義域為R,數(shù)列{an}滿足an=f(an-1)(n∈N*且n≥2).
(Ⅰ)若數(shù)列{an}是等差數(shù)列,a1≠a2,且f(an)-f(an-1)=k(an-an-1)(k為非零常數(shù),n∈N*且n≥2),求k的值;
(Ⅱ)若f(x)=kx(k>1),a1=2,bn=lnan(n∈N*),數(shù)列{bn}的前n項和為Sn,對于給定的正整數(shù)m,如果
的值與n無關(guān),求k的值.
(Ⅰ)若數(shù)列{an}是等差數(shù)列,a1≠a2,且f(an)-f(an-1)=k(an-an-1)(k為非零常數(shù),n∈N*且n≥2),求k的值;
(Ⅱ)若f(x)=kx(k>1),a1=2,bn=lnan(n∈N*),數(shù)列{bn}的前n項和為Sn,對于給定的正整數(shù)m,如果
S(m+1)n |
Smn |
數(shù)學(xué)人氣:775 ℃時間:2019-10-19 01:16:01
優(yōu)質(zhì)解答
(本小題共13分)(Ⅰ)當(dāng)n≥2時,因為an=f(an-1),f(an)-f(an-1)=k(an-an-1),所以an+1-an=f(an)-f(an-1)=k(an-an-1).因為數(shù)列{an}是等差數(shù)列,所以an+1-an=an-an-1.因為 an+1-an=k(an-an-1...
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