lga1+lga2+...+lgan=lg(a1·a2·a3····an)=n^2+n=>a1·a2·a3····an=e^(n^2+n)
所以a1·a2·a3····a(n-1)= e^((n-1)^2+n-1)
上下两式相除得 an=[e^(n^2+n)]/[e^((n-1)^2+n-1)] =e^(2n)(2)设bn=logan10*loga(n+1)10,数列{bn}的前n项和为Tn，证明 1/8≤Tn＜1/4logan10是以an 为底吗