数列an是公差不为零的等差数列求和公式
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(Ⅰ)设公差为d,由条件得5a1+5×42d=30(a1+2d)2=a1(a1+8d),得a1=d=2.∴an=2n,Sn=2n+n(n-1)×22=n2+n;(Ⅱ)∵1Sn+an+2=1n2+n+2
an=a1+(n-1)d=2+(n-1)da2=2+da4=2+3da8=2+7da2,a4,a8成等比数列,即a4/a2=a8/a4a4*a4=a2*a84+12d+9d^2=4+16d+7d^22
(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
2^n等比数列啊Sn=2+2^2+.+2^n2Sn=2^2+.+2^(n+1)2Sn-Sn=2^(n+1)-2Sn=2^(n+1)-2
a3=b3a1+2d=b1*q^2=a1*q^2a1+2d=a1*q^2.1a7=b5a1+6d=b1*q^4=a1*q^4a1+6d=a1*q^4.21式×3-2式2a1=3a1*q^2-a1q^4
2a3+2a11=4a7令a7=x-a7²+4a7=0解的a7=0(舍去),a7=4b6b8=b7²因为b7=a7结果为16再问:a7为什么不为0?再答:因为b7不能为0
(1)∵数列{an}是公差不为零的等差数列,a1=2,且a2,a4,a8成等比数列,∴(2+3d)2=(2+d)(2+7d),解得d=2,∴an=2n.(2)∵an=2n,∴3an=32n=9n,此数
1.设数列{an}的公差是d,则a(n+1)cosA+an*sinA=(an+d)*cosA+an*sinA=1即(cosA+sinA)*an=1-dcosA若cosA+sinA不等于0,则an=(1
用求和公式,求解二元二次方程组.
由题意,显然该等比数列的公比不会是负数,也不会是小于一的数.前者不会满足等差数列要求,后者末项趋于零,不合理.故公比大于一,故等差数列是递增的即公差大于0.又a5*a5=a3*an1即36=a3*an
设an=a+d*(n-1)1.a3+a10=a+2d+a+9d=2a+11d=152.a3*a7=a4*a4(a+2d)(a+6d)=(a+3d)^2a=-1.5d联立1与2,求得d=15/8a=-4
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
在等差数列中,公差d不为0,a11+40d=a51,即a11=a51-40d因为|a11|=|a51|,即a11=-a51,或者a11=a51(不符,舍去)所以a11+a51=2*a31=0,即a31
1Ab1·Ab3=Ab2^2就是A1·A17=A5^2则A1·(A1+16d)=(A1+4d)^2A1=2d那么q=A5/A1=(A1+4d)/A1=(2d+4d)/2d=3;q=32Abn=A1·3
(1)因为a4,a5,a8成等比数列,所以a52=a4a8.设数列{an}的公差为d,则(3+3d)2=(3+2d)(3+6d)化简整理得d2+2d=0.∵d≠0,∴d=-2.于是an=a2+(n-2
解a1=1a2=1+da5=1+4da1a2a5成等比所以(1+d)^2=1*(1+4d)d^2-2d=0d=2d=0(舍)所以an=a1+(n-1)d=1+(n-1)*2=2n-1
设公差为d,则a2=1+d,a5=1+4d,则1×(1+4d)=(1+d)2,∴d=2,∴an=2n-1,故答案为:2n-1.
a1a2a3成等比数列a2^2=a1a3=a3(a1+d)^2=a1+2da1^2+2a1d+d^2=a1+2d1+2d+d^2=1+2dd^2=0d=0公差不为零的等差数列错题
数列{an}是公差不为0的等差数列,设公差为d,S1,S2,S4成等比数列,则S22=S1•S4,∴( 2a1+d)2=a1•(4a1+6d),化简可得d=2a1∴a3a1=a1+2da1=
设首项为a1,公差为d.由题得:a1+a5=2*4a1*a7=a₃^2则:a1+(a1+4d)=8a1(a1+6d)=(a1+2d)^2综上解得a1=2d=1所以S5=20