$\begin{array}{l} A = {1^2} + {2^2} + {3^2} + ... + {101^2}\\ = 1 + 2\left( {1 + 1} \right) + 3\left( {2 + 1} \right) + ... + 101\left( {100 + 1} \right)\\ = \left( {1 + 2 + 3 + ... + 101} \right) + \underbrace {\left( {1.2 + 2.3 + 3.4 + ... + 100.101} \right)}_B\\ = \dfrac{{101.102}}{2} + B\\ \Rightarrow 3B = 1.2.3 + 2.3.3 + 3.3.4 + ... + 100.101.3\\ \Rightarrow 3B = 1.2.3 + 2.3.\left( {4 - 1} \right) + 3.4.\left( {5 - 2} \right) + ... + 100.101.\left( {102 - 99} \right)\\ \Rightarrow 3B = 100.101.102 \Rightarrow B = \dfrac{{100.101.102}}{3}\\ \Rightarrow A = \dfrac{{101.102}}{2} + \dfrac{{100.101.102}}{3} = \dfrac{{101.102.\left( {2.101 + 1} \right)}}{6}\\ \Rightarrow A = 348551 \end{array}$
