且n1,则logn(n+1)与log(n+1)(n+2)的大小关系为?
若n是正整数,且n>1,则logn(n+1)与log(n+1)(n+2)的大小关系为?
因为log(n+1)[n+2]+log(n+1)n=log(n+1)(n^2+2n)=2 所以 log(n+1)(n+2)
log(n+1)(n+2)-log(n)(n+1) (换底,使底数为n+1,并且予以省略) =log(n+2)-1/logn =[lognlog(n+2)-1]/logn 因为lognlog(n+2)0 故log(n+2)log(n+1)(n+2).
a=logn(n+1),n^a=n+1 b=log(n+1)(n+2),(n+1)^b=n+2, n^a+1=(n+1)^b>n^b+1^b=n^b+1 n^a>n^b,a>b logn(n+1)>log(n+1)(n+2)
loga[b]中a为底数. 1.logn[n+1]=1/log(n+1)[n], 2.log(n+1)[n+2]= =log(n+1)[n+1]+log(n+1)[(n+2)/(n+1)]= =1+log(n+1)[(n+2)/(n+1)]≤ ≤1+log(n+1)[(n+1)/n] 3.{log(n+1)[n+2]}/{logn[n+1]}= ={log(n+1)[n+2]}{log(n+1)[n]}= ={log(n+1)[n+2]}{log(n+1)[n+1]+log(n+1)[n/(n+1)]}= ={log(n+1)[n+2]}{1-log(n+1)[(n+1)/n]}≤ ≤{1+log(n+1)[(n+1)/n]}{1-log(n+1)[(n+1)/n]}= =1-{log(n+1)[(n+1)/n]}^2 log(n+1)[n+2]
解: n是正整数且n>1 所以 n(n+1)〈(n+1)(n+2) 一,当log底数a范围 01的时候 logn(n+1)〈log(n+1)(n+2) 三,挡当log底数a范围 a=1得时候,不成立 因为1得任何次方都是1
答:任意正整数n,最小是1. n=1,2^(n-1)=2^0=1,(n+1)^2=2^2=4; n=2,2^(n-1)=2^1=2,(n+1)^2=3^2=9; n...详情>>