已知等比数列{an}满足an>0,n=1,2,…,且a5-a2n-5=2^2n(n>=3),则当n>=1时,log2a1+log2a3+…log2a2n-1= 此题的答案是∵a[n]是等比数列.又有a[5]*a[2n-5]=2^2n ∴a[1]*a[2n-1]=2^2n设a[n]得公比为q.∵a[1]*q^n-1=a[n],a[2n-1]=a[1]*q^
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![已知等比数列{an}满足an>0,n=1,2,…,且a5-a2n-5=2^2n(n>=3),则当n>=1时,log2a1+log2a3+…log2a2n-1= 此题的答案是∵a[n]是等比数列.又有a[5]*a[2n-5]=2^2n ∴a[1]*a[2n-1]=2^2n设a[n]得公比为q.∵a[1]*q^n-1=a[n],a[2n-1]=a[1]*q^](/uploads/image/z/1358509-13-9.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BD%9Ban%7D%E6%BB%A1%E8%B6%B3an%3E0%2Cn%3D1%2C2%2C%E2%80%A6%2C%E4%B8%94a5-a2n-5%3D2%5E2n%28n%3E%3D3%29%2C%E5%88%99%E5%BD%93n%3E%3D1%E6%97%B6%2Clog2a1%2Blog2a3%2B%E2%80%A6log2a2n-1%3D+%E6%AD%A4%E9%A2%98%E7%9A%84%E7%AD%94%E6%A1%88%E6%98%AF%E2%88%B5a%5Bn%5D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97.%E5%8F%88%E6%9C%89a%5B5%5D%2Aa%5B2n-5%5D%3D2%5E2n+%E2%88%B4a%5B1%5D%2Aa%5B2n-1%5D%3D2%5E2n%E8%AE%BEa%5Bn%5D%E5%BE%97%E5%85%AC%E6%AF%94%E4%B8%BAq.%E2%88%B5a%5B1%5D%2Aq%5En-1%3Da%5Bn%5D%2Ca%5B2n-1%5D%3Da%5B1%5D%2Aq%5E)
已知等比数列{an}满足an>0,n=1,2,…,且a5-a2n-5=2^2n(n>=3),则当n>=1时,log2a1+log2a3+…log2a2n-1= 此题的答案是∵a[n]是等比数列.又有a[5]*a[2n-5]=2^2n ∴a[1]*a[2n-1]=2^2n设a[n]得公比为q.∵a[1]*q^n-1=a[n],a[2n-1]=a[1]*q^
已知等比数列{an}满足an>0,n=1,2,…,且a5-a2n-5=2^2n(n>=3),则当n>=1时,log2a1+log2a3+…log2a2n-1=
此题的答案是∵a[n]是等比数列.又有a[5]*a[2n-5]=2^2n ∴a[1]*a[2n-1]=2^2n
设a[n]得公比为q.∵a[1]*q^n-1=a[n],a[2n-1]=a[1]*q^2n-2,
∴a[1]*a[2n-1]=(a[1]*q*n-1)²=a[n]² ∴a[n]²=2^2n 从而 a[n]=2^n
∵log2 a[1]+log2 a[2]+…+log2a[2n-1]=log2(a[1]*a[2]*…a[2n-1])
=log2[(a[1]*a[2n-1])*(a[2]*a[2n-2])…
=log2[2^n²]
=n^2
(1)此题中n大于等于3.还有n大于等于1起到了什么作用?
(2)原题中写的是log2a1+log2a3+…log2a2n-1= ,怎么在做题时却出现了log2 a[2]+呢?
(3)∵A5*A2n-5=2^2n ∴A1*A2n-1=2^2n由上一步怎么得出的下一步呢?
已知等比数列{an}满足an>0,n=1,2,…,且a5-a2n-5=2^2n(n>=3),则当n>=1时,log2a1+log2a3+…log2a2n-1= 此题的答案是∵a[n]是等比数列.又有a[5]*a[2n-5]=2^2n ∴a[1]*a[2n-1]=2^2n设a[n]得公比为q.∵a[1]*q^n-1=a[n],a[2n-1]=a[1]*q^
回答:(1)n大于等于1的作用是:log2a(2n--1)有意义.
(2)题中的log2a(1)+log2a(2)+.+log2a(2n--1)=log2(a1)*a(2).*a(2n--1)
是运用了对数的性质“对数的和等于积的对数.”如“logaM+logaN=log a(M*N).”
(3) A5*A2n--5=A1*A2n--1.这是等比数列的基本性质.
例如 a1*a5=a2*a4,a6*a10=a7*a9=a2*a14等等.
懂了吗?