熱心網(wǎng)友

1)f(x)=log(2)x+log(4)x=logx+2/logx(省略底數(shù)2)f(2^an)=2n---log(2^an)+2/log(2^an)=2n---an+2/an=2n---(an)^2-2nan+2=0---an=n+√(n^2-2)2)a(n+1)-an=(n+1)+√[(n+1)^2-2]-[n+√(n^2-2)]=1+√[(n+1)^2-2]-√(n^2-2) 分子有理化得到:=1+(2n+1)/{√[(n+1)^2-2]+√(n^2-2)}當(dāng)n=2時此差顯然是正數(shù),所以a(n+1)an