等差数列{an}的前三项分别为a-1,a 1,2a-3则该数列的通项是
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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
An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn
由题意可得a1b1=S1T1=524=13,故a1=13b1.设等差数列{an}和{bn}的公差分别为d1 和d2,由S2T2=a1+a1+d 1b1+b1 +d&nbs
由AnBn=7n+45n+3,可设An=kn(7n+45)⇒an=An-An-1=14kn+38k,设Bn=kn(n-3)⇒bn=Bn-Bn-1=2kn+2k,所以a2n=28kn+38k,a2nbn
∵{an}为等差数列,其前n项之和为Sn,∴S2n-1=(2n−1)(a1+a2n−1)2=(2n−1)×2an2=(2n-1)•an,同理可得,S′2n-1=(2n-1)•bn,∴anbn=S2n−
(1)由于前三项之积为512所以:(a1)(a2)(a3)=(a2/q)(a2)(a2q)=(a2)³=512因此:a(2)=8且:a(1)-1,a(2)-3,a(3)-9成等差数列:\x0
答:1设an,bn的公差分别为d1,d2,Sn=na1+n(n-1)d1/2,Tn=nb1+n(n-1)d2/2,令S(n+3)=(n+3)a1+(n+3)(n+2)d1/2=Tn=nb1+n(n-1
题目错了!正确题目:已知递增的等比数列前三项之(积)为512,且这三项分别减去1,3,9 后又成等差数列,求证1/a1+2/a2+3/a3.+n/an<1/2 是不是?
由b+1-(b-1)=2b+3-(b+1)得b=0a8=-1公差d=2an=a8+(n-8)*d=2n-17
∵SnTn=2n3n+1,∴anbn=a1+a2n−1b1+b2n−1=S2n−1T2n−1=2(2n−1)3(2n−1)+1=2n−13n−1∴limn→∞anbn=limn→∞2n−13n−1=l
An=4n-3=a1+(n-1)dd=4a1=1等比数列{An}中,A5=7,A6=21,a8=(a6)^2/a5=63
∵等差数列{an}{bn}的前n项和分别为Sn,Tn,∵SnTn=7nn+3,∴a5b5=s9T9=7×99+3=6312=214,故答案为:214
a1=-2d=-3
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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
{an+qbn}(q为常数)的公差为:d1+qd2an=a1+(n-1)d1bn=b1+(n-1)d2,qbn=qb1+(n-1)qb2令cn=an+qbn则,cn=a1+qb1+(n-1)(d1+q
an+bn-(an-1+bn-1)=(an-an-1)+(bn-bn-1)=d1+d2,所以{an+bn}是等差数列,公差是d1+d2