[2%] Consider an algorithm with a running time function T(n)=50n2+200nlogn+106T(n) = 50n^2 + 200n\log n + 10^6T(n)=50n2+200nlogn+106. Is it mathematically correct to claim that T(n)=O(n2)T(n) = O(n^2)T(n)=O(n2)?