已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.(求通项)
来源:学生作业帮助网 编辑:六六作业网 时间:2024/10/06 12:53:34
已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.(求通项)已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.(求通项)已知
已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.(求通项)
已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.(求通项)
已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.(求通项)
a(n+1)=an+1+2√(an+1)
a(n+1)+1=(√(an+1)+1)^2
{√[a(n+1)+1]}^2=(√(an+1)+1)^2
{√[a(n+1)+1+√(an+1)+1)}{√[a(n+1)+1-√(an+1)-1)}=0
√(n+1)+1+√(an+1)+1)=0 或 √[a(n+1)+1-√(an+1)-1]=0
√[a(n+1)+1-√[(an+1]-1]=0
√[a(n+1)+1-√[(an+1=1
{√(an+1}成等差
√(an+1)=√(a1+1)+(n-1)=1+n-1=n
an=n^2-1
由a(n+1)=an+1+2根号下(an+1)得,a(n+1)+1=[根号下(an+1)+1]²。
而a(n+1)+1可以写成 根号下[a(n+1)+1]²,所以根号下a(n+1)+1=根号下(an+1)+1
也就是说 根号下(an+1) 成等差数列。
根号下(a1+1)=1,所以 根号下(an+1)=n,
∴an=n²-1
已知数列{an}满足a(n+1)=an+n,a1=1,则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a(n+1)=an+lg2,a1=1,求an
已知数列{an}满足a1=0,a(n+1)=an+2n,那么a2005的值是?
已知数列an中满足a1=1且当n.=2时,2an*a*(n-1)+an-a(n-1)=0,求通项公式an
已知数列{an}满足a1=1,3a(n+1)+an-7
已知数列an满足a1=1,a(n+1)=an/(3an+1) 求数列通项公式
已知数列{an}满足a1=33,a(n+1)-an=2n,则an/n的最小值
已知数列{an}满足a1=33,a(n+1)-an=2n,求an/n的最小值
已知数列an满足:a1=1,an-a(n-1)=n n大于等于2 求an
已知数列an满足a1=100,a(n+1)-an=2n,则(an)/n的最小值为
已知数列an满足a1=2,an=a(n-1)+2n,(n≥2),求an
已知数列{an}满足an>0,a1=3 根号下[a(n+1)]=(根号下an)+1 (n属于N*) 则an=?如题.
已知数列满足a1=1,an-a(n-1)=n-1,求其通项
已知数列{an}满足a1=1,a(n+1)=nan n+1是角标
已知数列{an}满足a1=2,a(n+1)-an=a(n+1)*an,则a31=?
已知数列{a}满足a1=1/2,a(n+1)=an+1/(n^2+n),求an已知数列{a}满足a1=1/2,a(n+1)=an+1/(n^2+n),求an
已知数列{an}满足a1=1,a(n+1)=an+√((an)^2+1),令an=tanθn(0