已知数列{an}满足a1=1,a2=-13,an+2-2an+1+an=2n-6b1=a2-a1=-13-1=-14bn=a(n+1)-ana(n+2)-2a(n+1)+an=2n-6a(n+2)-a(n+1)-a(n+1)+an=2n-6a(n+2)-a(n+1)-[a(n+1)+an]=2n-6b(n+1)-bn=2n-6所以bn-b(n-1)=2(n-1)-6bn-b(n-1)=2(n-1)-6.b3-b2=2*2-6b2-b1=2*1-6以上等
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/13 03:00:25
xSJ1똦CnQ|@(JVDAQ">PW*>>k&3_NiG[EW"{9sxj>Egც
Y g8 rB+#Brtg\ Ei}E2ހ
$&or䢩f`pz5^"LA54`(^Auڎꞿ0@ƈd[4%K:5ќ zlW^M&9#t<4Q>Wc-h=ɓ1!fMNSRi5eYr L58;lLJ:AkQ&ܭOBV[aN?oz9/s=Ǘ16my{n?83
ZB&qT;fghpz67+e!KXO~t֫%H\·S|]>
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
已知数列{an}满足关系式lg(1+a1+a2+.+an)=n,求数列{an}的通项公式
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
已知数列an满足a1=1,a2=3,an+1.an-1=an,求a2013
已知数列{an}满足关系式lg(1+a1+a2+.+an)=n,求数列的通项公式
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}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