设an是公差为正数的等差数列
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设根号Sn=d*n+HSn=d^2*n^2+2*d*H*n+H^2a1=S1=d^2+2*d*H+H^2a2=S2-S1=3*d^2+2*d*Ha3=S3-S2=5*d^2+2*d*Ha1+a3=2*
这是今年江苏卷上的题目…………(1)设根号Sn=d*n+HSn=d^2*n^2+2*d*H*n+H^2a1=S1=d^2+2*d*H+H^2a2=S2-S1=3*d^2+2*d*Ha3=S3-S2=5
1、√S1=√a1√S2=√(a1+a2)=√a1+2(1)√S3=√(a1+a2+a3)=√(3a2)=√a1+4(2)由(1)得a1+a2=a1+4√a1+4√a1=(a2-4)/4代入(2)√(
设根号sn=x+2(n-1)sn=(x+2(n-1))²an=sn-s(n-1)an=(x+2(n-1))²-(x+2(n-2))²an=(2x+2(n-1)+2(n-2
√Sn=√S1+(n-1)d√S2=√S1+d√S3=√S1+2d第2个式子两边平方a1+a2=a1+2(√a1)d+d^2第3个式子两边平方a1+a2+a3=a1+4(√a1)d+4d^2两个式子相
s1=a1;s2=a1+a2;s3=a1+a2+a3=3a2根号s3=根号s1+2d=根号s2+d化简得a2=3a1代入等差数列可求得d=根号a1sn=(nd)²an=sn-sn-1=(2n
结果是an=4(2n+1);首先由s1,s2,s3的关系可列出两个方程,关于a1,a2,a3.和已知的2a2=a1+a3联立,求出a1=4.接下来,利用根号sn是等差数列,推导出s(n)和a1的关系,
(a1)(b1)=1,因b1=1,则:a1=1则:(a2)(b2)=(a1+d)[b1q]=(1+d)q=4,则:(1+d)²q²=16(a3)(b3)=(a1+2d)[b1q
(1)根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项:an=n(2)根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0
a2+a3+a4=153*a3=15a3=5a2+a4=10(a3-1)的平方=a2*a4a2*a4=16可求a2=2a4=8或a2=8a4=2所以d=3或-3(舍)a1=a2-d=-1an=3n-4
1.{An}为等差数列,所以2a3=a2+a4a2+a3+a4=15=3a3a3=5(5-1)^2=16=(5-d)(5+d)d=3(-3舍掉)所以a1=5-2d=-1d=32.an=a1+(n-1)
a(1)*a(3)=a(2)^2代入解二次方程得a(2)=4舍掉不为正的解所以a(n)=2^nb(n)=2n-1∑an+∑bn=2^(n+1)-1+n^2再问:最后一步是什么意思,怎么得出来再答:求a
105a1+a2+a3=15,即3(a1+k)=15,a1+k=5,即a2=5a1*a2*a3=80,a1*a3=16,即a1=2,a3=8,k=3a11+a12+a13=a1+a2+a3+30k=1
(I)由a1,a2,a4成等比数列可得:(a1+2)2=a1(6+a1)∴4=2a1即a1=2∴an=2+2(n-1)=2n(II)∵bn=n•2an,=n•22n=n•4n∴Sn=1•4+2•42+
设该等差数列是首项为a1,公差为dS3=3a1+3(3-1)*d/2=3a1+3dS2=2a1+2(2-1)*d/2=2a1+dS4=4a1+4(4-1)*d/2=4a1+6d又:S3²=9
设公差=mA1=A2-mA3=A2+m所以3*A2=15A2=5所以(5-m)*5*(5+m)=80m=+3(-3舍去)所以A1=2,A2=5,A3=8
a1+a2+a3=3a2=15a2=5a1+a3=10a1*a3=16由此可得a1=2,a3=8(因为公差为正数,故a1
a2=5a1=2a3=8an=3n-1
再问:亲,若复数z满足iz=2,其中i为虚数单位,则z等于?再答:我们这边没学过复数。。再问:哦,也谢谢你的帮忙再答:不用谢~
根号Sn的通项公式是nSn=n^2an=Sn-Sn-1=n^2-(n-1)^2=2n-1