Question

数学，在线等！！！！快，详细过程

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

Answer 1

设公比为d，则
1/lga1*lga2+1/lga2*lga3+.....1/lgan-1*lgan

追问

then

追答

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

Answer 2

...........好深奥