Evaluate ∑n=160001n(n+1).\sum_{n=1}^{6000}\frac1{n(n+1)}.n=1∑6000n(n+1)1. A60006001\frac{6000}{6001}60016000B60016000\frac{6001}{6000}60006001C59996000\frac{5999}{6000}60005999D60005999\frac{6000}{5999}59996000E59996001\frac{5999}{6001}60015999