大师作文网设a1,a2,a3,.an都是正数,且构成等比数列,求证1/lga1*lga2+1/lga2*lga3+.1/lgan-
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设a1,a2,a3,.an都是正数,且构成等比数列,求证1/lga1*lga2+1/lga2*lga3+.1/lgan-1*lgan=n-1/lga1*lgan
设公比为d,则
1/lga1*lga2+1/lga2*lga3+.1/lgan-1*lgan
再问： then
再答： 1/lga1-1/lga2=(lga2-lga1)/(lg1*lga2)=lg(a2/a1)/(lga1*lga2)=lgd/(lga1*lga2) 1/lga2-1/lga3=(lga3-lga2)/(lg2*lga3)=lg(a3/a2)/(lga2*lga3)=lgd/(lga2*lga3) ... d不等于1时： 1/lga1*lga2+1/lga2*lga3+.....1/lgan-1*lgan =(1/lgd)[(1/lga1-1/lga2)+/(1/lga2-1/lga3)+...+(1/lga(n-1)-1/lgan) =(1/lgd)[1/lga1-1/lgan] =(1/lgd)[(lgan-lga1)/(lga1lgan)] =(1/lgd)[(n-1)lgd/(lga1lgan)] =(n-1)/(lga1lgan) d=1时显然
1/lga1*lga2+1/lga2*lga3+.1/lgan-1*lgan
再问： then
再答： 1/lga1-1/lga2=(lga2-lga1)/(lg1*lga2)=lg(a2/a1)/(lga1*lga2)=lgd/(lga1*lga2) 1/lga2-1/lga3=(lga3-lga2)/(lg2*lga3)=lg(a3/a2)/(lga2*lga3)=lgd/(lga2*lga3) ... d不等于1时： 1/lga1*lga2+1/lga2*lga3+.....1/lgan-1*lgan =(1/lgd)[(1/lga1-1/lga2)+/(1/lga2-1/lga3)+...+(1/lga(n-1)-1/lgan) =(1/lgd)[1/lga1-1/lgan] =(1/lgd)[(lgan-lga1)/(lga1lgan)] =(1/lgd)[(n-1)lgd/(lga1lgan)] =(n-1)/(lga1lgan) d=1时显然
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