已知数列{an}满足a1=1,a2=2,a(n+2)=a(n+1)-an,则a100=____________
来源:学生作业帮助网 编辑:作业帮 时间:2024/12/02 19:09:59
xSn0<[yTJ+|@T#B*h[3- 6RjSYDi!v'~c;)TM<,j{=\p?Su+*}VBD7Uw]vuQ
4pLBVl>v6i6xpC=nɡMT'Md/]}pDWGpamW^β,=y'q<4w_,XdzRY-K^I4J@;q{Ȕ"?'i\dCܧ<ѧc%˧g2Dei8AJOQޯJ:V=ukYP\\MG:}!oo,5~Ep5Ҁ&{QܦHFe6+FEu>2?n
]Q^mq;T2PMxwe6ѥ:?Խ. KX~}-={N[_~0
已知数列{an}满足a1=1,a2=2,a(n+2)=a(n+1)-an,则a100=____________
已知数列{an}满足a1=1,a2=2,a(n+2)=a(n+1)-an,则a100=____________
已知数列{an}满足a1=1,a2=2,a(n+2)=a(n+1)-an,则a100=____________
a(n+2)=a(n+1)-an
a(n+1)=an -a(n-1)
所以
a(n+2) = -a(n-1)
an = -a(n-3) = a(n-6)
a(100) = a(100 - 16*6) = a4 = -a1 = -1
别只出题,不处理题哟.该采纳一些了.
4
这个数列应该为
1,2,1,-1,2,3,1,-2,-3,-1,2,3,1,-2,-3,
,-1,2,3,1,-2,-3,六个数循环
(100-3)/6=16余1
答案为-1
6个一周期
答案为-1
a(100=)a(98+2)=a(98+1)-a(98)=a(99)-a(98)-----1式,以此类推。a(99)=a(98)-a(97)------2式,a(99)=a(97)-a(96),1式减2式,以此类推,a(100)=a(2)-a(1)=2-1=1
1,2,1,-1,-2,-1,1,2,1,-1,-2,-1....
可见数列以每6个数为一个周期
100是除以6的余数为4,故,a100的值为6个数一个周期的第四个数
即a100=-1
-1
已知数列an满足an=1+2+...+n,且1/a1+1/a2+...+1/an
已知数列an'满足a1=1/2,a1+a2+a3+...+an=n^2an,求通项公式
数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an
已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an求an
已知数列满足a1=1/2,an+1=2an/(an+1),求a1,a2已知数列满足a1=1/2,a(n+1)=2an/(an+1),求a1,a2;证明0
已知数列an满足a1=0 a2=1 an=(An-1+An-2)/2 求liman
已知数列an满足a1=0 a2=1 an=(An-1+An-2)/2 求liman
几个数列问题.已知数列{an} a1=1,an+1=an/(1+n^2*an) 求an 已知数列{an} 满足a1=1 a1*a2*a3.*an=n^2 求an
已知数列an满足an=1+2+...n,且(1/a1)+(1/a2)+...(1/an)
已知数列{an}满足:a1+a2+a3+.+an=n^2,求数列{an}的通项an.
已知数列{an}中满足a1=1,a(n+1)=2an+1 (n∈N*),证明a1/a2+a2/a3+…+an/a(n+1)
已知数列{an}满足条件:a1=5,an=a1+a2+...a(n-1) n大于等于2,求数列{an}的通项公式
已知递增数列{an}满足a1=1,(2an+1)=an+(an+2),且a1,a2,a4成等比数列.求an
关于数列极限的已知数列an满足a1=0 a2=1 an=(an-1+an-2)/2 求lim(n->无穷)an
已知数列{an}满足:a1=1,且an-an-1=2n,求(1)a2,a3,a4.(2)求数列{an}的通项an
已知数列(an)满足a1=1,an+1=2an/an+2(n∈N*) 求a2,a3,a4,a5 猜想数列(an)的通项公
已知数列an中 a1=1a2=2
(1)数列{an}中,a1=1,a2=-3,a(n+1)=an+a(n+2),则a2005=____(2)已知数列{an}满足a1=1,a1×a2×a3…an=n^2,求an.