a(n+1)-a(n)=k(a(n)-a(n-1)),a(n+1)-a(n)為等比數(shù)列,公比為k,首項(xiàng)為a2-a1
a(n)等差,a(n+1)-a(n)=a(n)-a(n-1)=d,k=1
已知定義在R上的函數(shù)f(x)和數(shù)列{an},a1=a,a2不等于a1,當(dāng)n屬于N*且n大于等于2時(shí),an=f(an-1),且
已知定義在R上的函數(shù)f(x)和數(shù)列{an},a1=a,a2不等于a1,當(dāng)n屬于N*且n大于等于2時(shí),an=f(an-1),且
f(an)-f(an-1)=k(an-an-1),其中a,k為非零常數(shù)
(1),若數(shù)列an是等差數(shù)列,求k值
f(an)-f(an-1)=k(an-an-1),其中a,k為非零常數(shù)
(1),若數(shù)列an是等差數(shù)列,求k值
數(shù)學(xué)人氣:427 ℃時(shí)間:2019-08-18 18:09:46
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