a1=1,a2=5,a3=17,a4=53则数列an的通项公式为什么?
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lingo MODEL:sets:banci/1..12/:a1,a2,a3,a4,a5,b;endsetsmin=z;z=@smax(a1(1)+a2(1)+a3(1)+a4(1)+a5(1),a1(2)+a2(2)+a3(2)+a4(2)+a5(2),a1(2)+a2(2)+a3(2)+a4(2)+a5(2),a1(3)+a2(3)+a3(3)+a4(3)+a5(3),a1(4)+a2(4)+a3(4)+a4(4)+a5(4),a1(5)+a2(5)+a3(5)+a4(5)+a5(5),a1
已知(x-1)^5=a5x^5+a4^4+a3^3+a2^2+a1^1+a0,则a5+a4+a3+a2+a1+a0=?,-a5+a4-a3+a2-a1+a0=?,a4+a2a4+a2=?
(a1+a3+a9)/(a2+a4+a10)=?
等差数列中a1+a2+a3+a4+a5=3,a1*a1+a2*a2+a3*a3+a4*a4+a5*a5=12求a1-a2+a3-a4+a5
已知a1,a2,a3,a4,a5是5个整数且a1=1,a5=6,求证a2-a1,a3-a2,a4-a3,a5-a4中至少有一个大于等于2
(a1+a2+a3+a4)/(a1+a2)=5 (a1+a2+a1*q^2+a2*q^2)/(a1(a1+a2+a3+a4)/(a1+a2)=5(a1+a2+a1*q^2+a2*q^2)/(a1+a2)=5 得q^2+1=5 这里看不懂
线代证明题证明:设有向量组a1,a2,a3,a4,若R(a1,a2,a3,a4)>R(a1,a2,a3)则必有R(a1,a2,a3,a4)=R(a1,a2,a3)+1
已知R(a1,a2,a3)=3,R(a1,a2,a3,a4)=3,R(a2,a3,a4)=2.证明:1)a4能由a2,a3,线性表示;2)a1不能由a2,a3,a4表示.
已知A={a1,a2,a3,a4,a5},B={a1^2,a2^2,a3^2,a4^2,a5^2},ai属于N*,i=1,2,3,4,5,设a1<a2<a3<a4<a5,且A∩B={a1,a4},a1+a4=10,又AUB元素之和为224,求a1,a4,a5及集合A
计算 n+1阶行列式,Dn+1=[x a1 a2 a3...an;a1 x a2 a3...an;a1 a2 x a3...an;............a1 a2 a3 a4 ...x]
设A=(a1,a2,a3,a4),ai(i=1,2,3,4)为5维向量,若a2,a3,a4线性无关,且a4=a1+2a2-a3,求方程组Ax=0的通解
设A=(a1,a2,a3,a4),ai(i=1,2,3,4)为5维列向量,若a2,a3,a4线性无关,且a4=a1+2a2-a3,求方程组AX=0的通解
定义enum number{a1,a2,a3=5,a4},则a1,a2,a4的值分别是多少
a1、a2、a3、a4为列向量若|a1 a2 a3|=3,|a4 a2 a1|=2,则|a1+a4+a3 a1 a2|=
a1a2a3a4三维列向量A=(a1,a2,2a3-a4+a2),B=(a3,a2,a1),C=(a1+2a2,2a2+3a4,a4+3a1),若|B|=-5,|C|=40,求|A|请尽快
设集合A=﹛a1,a2,a3,a4,a5},B={a1²,a2²,a3²,a4²,a5²}.设集合A=﹛a1,a2,a3,a4,a5},B={a1²,a2²,a3²,a4²,a5²},其中ai(1≤i≤5,i∈N)都是正整数,且a1<a2<a3<a4<a5,a1+a4=10,
等差数列 A1+A2+A3=6,A2+A3+A4=7,A3+A4+A5=
已知数列a1,a2,a3为等差数列,数列a2,a3,a4为等比数列,且a1+a4=16,a2+a3=12,求a1,a2,a3,a4=?