已知数列﹛an﹜,满足a1=4,a2=2,a3=1,数列﹛an+1-an﹜为等差数列,则an的通项公式为?an+1,指的是角标n+1
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 06:28:34
![已知数列﹛an﹜,满足a1=4,a2=2,a3=1,数列﹛an+1-an﹜为等差数列,则an的通项公式为?an+1,指的是角标n+1](/uploads/image/z/2735944-16-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%2C%E6%BB%A1%E8%B6%B3a1%3D4%2Ca2%3D2%2Ca3%3D1%2C%E6%95%B0%E5%88%97%EF%B9%9Ban%2B1-an%EF%B9%9C%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%88%99an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E4%B8%BA%3Fan%2B1%2C%E6%8C%87%E7%9A%84%E6%98%AF%E8%A7%92%E6%A0%87n%2B1)
x͒J1_%ˎIfHf,DH2PZAQ]P8&Ӯ
$3܊;sBOks'\`}<}yV"Od*ᡂ;[}'^czxvFqn!z63\
eo
ZDeɉ DTX* W:XZ[#hp)JM]LEm=A7#BށY]-Ζ'
KG`EM
x
H~s,sJ`@hwEBO4̢RAa=g
已知数列﹛an﹜,满足a1=4,a2=2,a3=1,数列﹛an+1-an﹜为等差数列,则an的通项公式为?an+1,指的是角标n+1
已知数列﹛an﹜,满足a1=4,a2=2,a3=1,数列﹛an+1-an﹜为等差数列,则an的通项公式为?
an+1,指的是角标n+1
已知数列﹛an﹜,满足a1=4,a2=2,a3=1,数列﹛an+1-an﹜为等差数列,则an的通项公式为?an+1,指的是角标n+1
bn = a(n-1) - an,
b1 = a2 -a1 = -2;
b2 = a3 - a2 =-1;
d = b2 -b1 = 1;
bn = a(n+1)-an
=b1 + (n-1)d
= -2 + n-1
= n -3;
a(n+1) - an = n-3;
an - a(n-1) = n-4;
...
a2 - a1 = -2;
an - a1 = -2+ -1 +...+ n-4
= (-2+n-4)*(n-1)/2
= (n-6)*(n-1)/2
an = (n-6)*(n-1)/2 + 4
= (n^2 - 7n +6)/2 + 4
= n^2/2 -7n/2 + 7.
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﹜,满足a1=4,a2=2,a3=1,数列﹛an+1-an﹜为等差数列,则an的通项公式为?an+1,指的是角标n+1
已知数列an'满足a1=1/2,a1+a2+a3+...+an=n^2an,求通项公式
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
数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an
已知数列{an}满足a1+a2+...+an=n²an求其通项an
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a1=4/3,2-a(n+1)=12/an+6则1/a1+2/a2+.1/an=?
已知数列an满足a1=1,an=a1+2a2+3a3+4a4+.(n-1)a(n-1),求通项an
已知数列{An}满足a1=1,a2=5,an+1=5an-4an-1,(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}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a1=½,a1+a2+.+an=n²an,求其通项an
已知数列{an}满足a1=½,a1+a2+.+an=n²an,求其通项an