问个问题,这个结论是怎么得到的? 1.T(n)=2T(n2)+nlogn1.T(n)=2T(\frac n2)+n\log n1.T(n)=2T(2n)+nlogn =4T(n4)+2×n2logn2+nlogn=4T(\frac n4)+2\times\frac n2\log \frac n2+n\log n=4T(4n)+2×2nlog2n+nlogn =8T(n8)+nlogn4+nlogn2+nlogn=8T({n\over8})+n\log{n\over4}+n\log{n\over2}+n\log n=8T(8n)+nlog4n+nlog2n+nlogn =…=\dots=… T(n)=∑i=0lognnlogn2iT(n)=\sum_{i=0}^{\log n}n\log\frac n{2^i}T(n)=∑i=0lognnlog2in =n∑i=0lognlogn2i=n\sum_{i=0}^{\log n}\log\frac n{2^i}=n∑i=0lognlog2in =n∑i=0lognlog2i=n\sum_{i=0}^{\log n}\log2^i=n∑i=0lognlog2i =n∑i=0logni=n\sum_{i=0}^{\log n}i=n∑i=0logni =Θ(nlog2n)=\Theta(n\log^2n)=Θ(nlog2n)
其他都懂,就是倒数第二行到倒数第一行不太懂,求助。