在等差数列an的前三项为a-1,1 2a 2
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![在等差数列an的前三项为a-1,1 2a 2](/uploads/image/f/3262949-53-9.jpg?t=%E5%9C%A8%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8D%E4%B8%89%E9%A1%B9%E4%B8%BAa-1%2C1+2a+2)
即对任意n∈N,(a+n)/(a+n-1)≥(a+8)/(a+7)两边同减1:1/(a+n-1)≥1/(a+7)此不等式可分三种情况:(1)a+7≥a+n-1〉0显然n≥8时不成立(2)0〉a+n-1
设公差为d,公比为q,则b2=qb1=q(a1+1)=(a1+d+2),↔2q=3+d,b3=q²b1=q²(a1+1)=(a1+2d+3),↔q²
已知等差数列{an}的前三项依次为a-1,a+1,2a+3,故有2(a+1)=a-1+2a+3,解得a=0,故等差数列{an}的前三项依次为-1,1,3,故数列是以-1为首项,以2为公差的等差数列,故
逆命题是:在公比不为1的等比数列{an}中,前n项的和为Sn,若a2,a4,a3成等差数列,则S2,S4,S3成等差数列.证明:设公比为q,则a2=a1q,a4=a1q³,a3=a1q&su
S2-S1=(an+1-a1)+(an+2-a2)+...+(a2n-an)=nd*n=d*n^2S3-S2=(a2n+1-a1)+(a2n+2-a2)+...+(a3n-a2n)=nd*n=d*n^
n=1时,a2-a1=3;n=2时,a3-a2=3+d;n=3时,a4-a3=3+2d;...n=n时,a(n+1)-an=3+(n-1)d;左右相加,得:a(n+1)-a1=3n+n*(n-1)d/
1.an=a1+(n-1)d=2+n-1=n+1Sn=(a1+an)*n/2=n(n+3)/22.bn=2^(n+1)bn是以b1=4为首项,2为公比的等比数列,Tn=b1(1-q^n)/(1-q)=
A1=a-1A2=a+2A3=2a+3D=(A3-A1)/(3-1)=A-1=(A2-A1)/(2-1)=3所以D=3A1=2AN=2+(N-1)3=3N-1
a8+a14=2a1+20d=0a1=-10d0Sn=na1+n(n-1)d/2=-10nd+n^2d/2-nd/2=(d/2)*n^2-(21d/2)n,对称轴是n=21/2=10.5所以,当n=1
等差数列{an}的前三项为a-1,a+1,2a+3所以:a-1+2a+3=a+1+a+13a+2=2a+2a=0所以前3项是-1,1,3an=-1+2(n-1)=2n-1Sn=-n+n(n-1)=n^
你可以看出公差d=2第一项是A-1所以公式为An=A1+(n-1)d即首项+(n-1)乘以公差d=a-1+(n-1)2=a+2n-3
2a-(a+1)=a+3-2a推导出啊a=2{an}=3,4,5...{an}=a+2(a>=1)
an=2n-3x+1-(x-1)=2x+3-(x+1)x=0d=2an=-1+(n-1)x2=2n-3
∵等差数列{an}的前三项分别为a-1,2a+1,a+7,∴2(2a+1)=a-1+a+7,解得a=2.∴a1=2-1=1,a2=2×2+1=5,a3=2+7=9,∴数列an是以1为首项,4为周期的等
(1)a(n+1)=an+2^n+1an=a(n-1)+2^(n-1)+1.a2=a1+2^1+1把上面n个等式相加得a(n+1)=2^n+2^(n-1)+.+2^1+n=2^(n+1)+n-2所以a
由题:a-1,a+1差为2,所以2a+3-(a+1)=2可得a=0;此等差数列为-1,1,3.通式为:2n-1,n=0,1,2.
d=2a+1-a=4a+2-(2a+1)a+1=2a+1得出a=0d=1第五项=4
设等差数列{an}的公差为d,(d>0)则1+2d=(1+d)2-4,即d2=4,解得d=2,或d=-2(舍去)故可得an=1+2(n-1)=2n-1,Sn=n(1+2n−1)2=n2,故答案为:2n
{an}是首项为a公差为1的等差数列,∴数列{an}的通项公式为an=a+n-1,∵bn=1+anan=1+1an=1+1a+n−1.∵bn≥b8∴1+1an≥1+1a8,即1an≥1a8,数列{an
1、a(n+1)=(2an)/(an+2)取倒数得:1/a(n+1)=(an+2)/(2an),1/a(n+1)=1/2+1/an,所以{1/an}是等差数列,公差是1/2,1/an=1+(n-1)*