有窮數(shù)列{an}}（n屬于N）滿足a1=an,a2=an-1,…,an=a1,即ai=an-i+1（n屬于N且1≤i≤n）.

有窮數(shù)列{an}}（n屬于N*）滿足a1=an,a2=an-1,…,an=a1,即ai=an-i+1（n屬于N*且1≤i≤n）.
有窮數(shù)列{an}（n屬于N*）滿足a1=an,a2=an-1,…,an=a1,即ai=an-i+1（n屬于N*且1≤i≤n）.若｛an｝的項(xiàng)數(shù)為7,且a1,a2,a3,a4成等差數(shù)列,a1=2,a4=11,試寫(xiě)出｛an｝的每一項(xiàng)

數(shù)學(xué)人氣：749 ℃時(shí)間：2020-05-15 03:55:28

優(yōu)質(zhì)解答

a(1)=a(7)=2,a(2)=a(6),a(3)=a(5),a(4)=11.a(2) = a(1) + d = 2+d,a(3) = a(1) + 2d = 2 + 2d,11 = a(4) = a(1) + 3d = 2 + 3d,d = 3.a(1)=a(7)=2,a(2)=a(6)=2+d=5,a(3)=a(5)=2+2d=8,a(4)=11n不是在【1，7】么，就這么簡(jiǎn)單？題目中說(shuō)“{a(n)}的項(xiàng)數(shù)為7” 啊。。也許是我想的太復(fù)雜了？3Q啦~多謝樓主的及時(shí)采納。。很高興有機(jī)會(huì)為樓主效勞~~

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有窮數(shù)列{an}}（n屬于N*）滿足a1=an,a2=an-1,…,an=a1,即ai=an-i+1（n屬于N*且1≤i≤n）.

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有窮數(shù)列{an}}（n屬于N）滿足a1=an,a2=an-1,…,an=a1,即ai=an-i+1（n屬于N且1≤i≤n）.