已知数列an为等比数列an大于0且a2a6 a4a5 a4a6=36
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解a(n+1)=pan+q这类题型常用方法如下:设a(n+1)+λ=μ(an+λ),然后求出λ、μ的值,即数列{an+λ}是等比数列设a(n+1)+λ=μ(an+λ),即a(n+1)=μan+μλ-λ
a(n+1)=2an/(an+1)∴1/a(n+1)=(an+1)/2an=1/2an+1/2∴1/a(n+1)-1=1/2an+1/2-1=1/2an-1/2=(1/2)(1/an-1),1/a1-
An=A1+(A2-A1)++(An-An-1)=1(1-(1/3)^n)/(1-1/3)=2/3-2(1/3)^(n+1)
an=1/5*(7/2)^(n-1)再问:过程啊再答:不好意思,答案应该是an=3^(n-1),刚才一不小心看错了现在这是对的,已经验算过了,不懂可以追问,祝愉快,望采纳。
a3^2+2a3*a5+a5^2=49(a3+a5)^2=49a3+a5=7再问:-7把再答:嗯忘看了an
设数列An的公比为q则:An=(a1)q^(n-1)而:a10^2=a15所以:((a1)q^(10-1))^2=(a1)q^(15-1)q^4=1/a1因q>1,因此:a1>0设另有数列Bn,Bn=
a1=1,a2=q,a3=q^2,则a1+a2+a3=1+q+q^2=7,即q^2+q-6=0,解得q=2或q=-3(舍去),所以q=2,所以an=a1×q^(n-1)=2^(n-1)
因为a1a2a3=8所以a2/q*a2*a2*q=8a2^3=8,a2=2又a1+a2+a3=7即a2/q+a2+a2*q=71/q+q=5/2=2+1/2所以q=2或1/2即a1=1或4.所以an=
a(n+1)+1/2=3an+1+1/2=3(an+1/2)a1+1/2=1所以{an+1/2}是以1为首相,3为公比的等比数列an+1/2=3^(n-1)an=3^(n-1)-1/2
设数列的公比为q,首项为a1,则∵a52=a10,2(an+an+2)=5an+1,∴(a1q4)2=a1q9,2(1+q2)=5q,∵等比数列{an}为递增数列,∴q=2,a1=2∴an=2n故答案
等比数列an的公比大于1,设公比为q,且q>1a1a3=6a2,a1*a2*q=6a2a1*q=6a2=6a1.a2.a3-8成等差,2a2=a1+a3-82*6=6/q+6*q-820q=6+6q^
不懂可以追问,一定尽力解答,祝愉快,望采纳.(图片点开就清晰了)再问:真的很感谢很认真详细再答:不客气,你懂了才是最重要的.再问:请问下午你会在吗有数学问题要问呢再答:不好意思,那天不在.现在还需要问
a(n+1)+1=2an+2=2(an+1)[a(n+1)+1]/(an+1)=2所以an+1是等比数列[a(n+1)+1]/(an+1)=2则q=2所以an+1=(a1+1)*2^(n-1)=2^n
n=b1.q^(n-1)bn=an-3nan=bn+3n=b1.q^(n-1)+3nSn=a1+a2+...+an=b1(q^n-1)/(q-1)+3n(n+1)/2
第一个晕,才看懂.明显公比是1第二个a3*a5=4提示你等比数列中,a2*a4=a3的平方a4*a6=a5的平方所以a3+a5平方=25an大于0,所以a3+a5=5,所以a3=1a5=4公比是2,a
an+Sn=2n令n=1a1+S1=2=>a1=1又a(n-1)+S(n-1)=2(n-1)与上式作差an-a(n-1)+an=22an-a(n-1)=2an-2=(1/2)[a(n-1)-2]得证a
a3+a4=a1q²+a1q³=q²(a1+a1q)=q²(a1+a2)=q²*3=12q=-2或者q=2当q=-2时a1+a2=a1+a1q=a1(
∵Sn=kq^n-k∴S(n+1)=kq^(n+1)-k∴a(n+1)=S(n+1)-Sn=[kq^(n+1)-k]-(kq^n-k)=k[q^(n+1)-q^n]=k[(q-1)q^na(n+1)/
若数列{lgan}为等差数列,可得:2lgan=lgan-1+lgan+1,即lgan2=lg(an-1•an+1),∴an2=an-1•an+1,∴数列{an}为等比数列;但数列{an}为等比数列,
a2=a1+da4=a1+3da2^2=a1a4a2^2=(a1+d)^2=a1^2+2a1d+d^2a1a4=a1(a1+3d)=a1^2+3a1da1^2+2a1d+d^2=a1^2+3a1da1