设数列{an}满足a1=a,a(n+1)=can+1-c,其中a,c为实数,且c不等于0,则数列{an}的通项公式a(n+1)=can+1-ca(n+1)-1=c(an-1)[a(n+1)-1]/(an-1)=c数列an-1等于(a-1)乘以c的n-1次方(我想知道这一步是怎么来的)an等于(a-1)乘
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/09 22:47:47
xSn@~z3^m.RiS/iU$Bk*I B*`PR .+tˏiMOU/eɞovfFI==n" zL$5vBD/~aaW$ŚhZ>Y}@rMlqKUM@$G~ƙ|`<(ᑥ7YgkR#L'ormYP&Xp~摾ɤ<{;ų6Zh?/A=y[D:ヲ-sKKU9|O"AoU'sAU@2<nG?7
ob2s;CW*}=s(OZҲFd偒xFzjA[Ƒ8|NHV6{>(̯GT9e>/VIrfcfA;6(L$?(w#YU~I/L)H,k7Æ7/ CrYA5&o%tIk̛Ta@"bv~I[Y,yd:VaĢ!Ubl5!Ş.#{˞}^{Z@>ؠ.Oq
T:^~Zek? t&
设数列an满足a1=2,a(n+1)=3an+2^(n-1),求an2,设数列an满足a1=2,a(n+1)=3an+2n,求an
设数列an满足a1=2,a(n+1)-an=3x2的2n-1次方,求数列an的通项公式
已知数列{an}满足a(n+1)=an+n,a1=1,则an=
数列{an}满足a1=2,a(n+1)=2an+n+2,求an
设数列an满足a1=1,a2=4,a3=9,an=a(n-1)+a(n-2)-a(n-3).则a2011=
数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010
已知数列{an}满足a(n+1)=an+lg2,a1=1,求an
数列[An]满足a1=2,a(n+1)=3an-2 求an
设数列{an}满足:存在正数M,对一切n有An=|a2-a1|+|a3-a2|+-----+|an-a(n-1)|
数列{an}中,a1=8,a4=2且满足a(n+2)=2a(n+1)-an,n属于N*数列{an}中,a1=8,a4=2且满足a(n+2)=2a(n+1)-an,n属于N*1.求数列{an}的通项公式2.设Sn=|a1|+|a2|+...+|an|,求Sn3.设bn=1/n(12-an)[n属于N*]是否存在最大的整数m,使得
设数列{an}满足a1+3a2+3的平方倍a3+..+3的n-1次方*an=n/3.a属于N*,设bn=n/an,求数列{bn}的前n项和Sn
已知数列{an}满足a1=1,an+1=2an+2.(1)设bn=2^n/an,求证:数列{bn}是等差数列.(2)求数列{an}的通项公式.a(n+1)
已知数列{an}满足A1=2,An+1=An - 1/n(n+1) (1)求数列an的通项公式 (2)设{Bn}=nAn*2^n,求数列Bn前n项和SnRT已知数列{an}满足A1=2,An+1=An - 1/n(n+1) (1)求数列an的通项公式(2)设{Bn}=nAn*2^n,求数列Bn前n项和Sn是A(n+1)
设数列{an}满足a1=a,an+1=can+1-c,n∈N*,其中a,c为实数,且c≠0,a≠11)求证{an-1}是等比数列2)求数列{an}的通项公式3)设a=1/2,c=1/2,bn=n(1-an),n∈N*,求证数列{bn}的前n和sn<2设数列{an}满足a1(第一项)=a,an+1(第n+1
设b>0,数列{An}满足A1=b,An=nbA(n-1)/A(n-1)+2n-2(n>=2).(1)求数列{An}的通项公式;(2)证明:对于一切正整数n,An
设数列{an}满足a1=2,a(n+1)-an=3乘以2的(2n-1}次方 1.求数列的通项公式; 2.令bn=n乘以an,求数列前n项和
已知数列{an}满足a1=31,a(n)=a(n-1)-2(n大于等于2,n属于自然数)设bn=|an|,求数列{an}的前n项和Tn
设数列an满足a1+3a2+3^2a3+……+3^(n-1)an=n/3,a是正整数,设bn=n/an,求数列bn的前n项和