若{an}为正项的等比数列,求证1/(lga1*lga2)+1/(lga2*lga3)+...+1/(lga(n-1)*lgan)=(n-1)/(lga1*lgan)

【证】设等比数列公比为q
1/(lga1*lga2)+1/(lga2*lga3)+...+1/(lga(n-1)*lgan)
=1/[lga1(lga1+lgq)]+1/[(lga1+lgq)(lga1+2lgq)]+……+1/[lga1+(n-2)q][lga1+(n-1)q]
={1/lga1-1/(lga1+lgq)+1/(lga1+lgq)-1/(lga1+2lgq)+……-1/[lga1+(n-1)q]}/lgq
={1/lga1-1/[lga1+(n-1)q]}/lgq
=(n-1)/(lga1*lgan)