已知数列{an}满足,a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,令bn=a(n+1)-an,证明:{bn}是等比数列(2)求{an}的通项
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/19 04:19:31
xRJA}/wE.%0PRH*\zpVWh5`/9yeY u #SF!> $~S8݆YT,jB%Ojq*EH;LڕNp1?8A:빳~U_h`)+
COB=:b(<V,D01tʐ0N
E>.6^MREW"[&@9Z)XųnSemZa=g5d#hX FG~2]l8k)lgA.#m?膓
已知数列{an}满足,a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,令bn=a(n+1)-an,证明:{bn}是等比数列(2)求{an}的通项
已知数列{an}满足,a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,令bn=a(n+1)-an,证明:{bn}是等比数列
(2)求{an}的通项
已知数列{an}满足,a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,令bn=a(n+1)-an,证明:{bn}是等比数列(2)求{an}的通项
证明 令bn=a(n+1)-an
2a(n+2)=an+a(n+1)
∴2[a(n+2)-a(n+1)]=an-a(n+1)=-[a(n+1)-an]
bn=a(n+1)-an,∴2b(n+1)=-bn,即b(n+1)/bn=-1/2
∴{bn}是等比数列
b1=a2-a1=2-1=1,{bn}是首项为1,公比为-1/2的等比数列
∴bn=1*(-1/2)^(n-1)
∴a(n+1)-an=(-1/2)^(n-1)
∴an-a(n-1)=(-1/2)^(n-2),
a(n-1)-a(n-1)=(-1/2)^(n-3)
……
a2-a1=(-1/2)^0
上面各式叠加得 an-a1=(-1/2)^0+……+(-1/2)^(n-3)+(-1/2)^(n-2)
=[1-(-1/2)^(n-1)]/(1+1/2)=(2/3)[1-(-1/2)^(n-1)]
∴an=a1+(2/3)[1-(-1/2)^(n-1)]=5/3-(2/3)*(-1/2)^(n-1)=5/3+(1/3)*(-1/2)^(n-2)
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.