Cho \(P = 2 \cdot 101 + 3 \cdot 100 + 4 \cdot 99 + ... + 99 \cdot 4 + 100 \cdot 3 + 101 \cdot 2\)

và \(Q = {2^2} + {3^2} + ... + {98^2} + {99^2} + {100^2} + {101^2}\).

Hãy tính tổng \(P + Q\).

        \( = \left( {2 \cdot 101 + {2^2}} \right) + \left( {3 \cdot 100 + {3^2}} \right) + \cdot \cdot \cdot + \left( {100 \cdot 3 + {{100}^2}} \right) + \left( {101 \cdot 2 + {{101}^2}} \right)\)

        \( = 2 \cdot \left( {101 + 2} \right) + 3 \cdot \left( {100 + 3} \right) + \cdot \cdot \cdot + 100 \cdot \left( {3 + 100} \right)\)

        \( = 2 \cdot 103 + 3 \cdot 103 + 4 \cdot 103 + ... + 101 \cdot 103\)

        \[ = 103 \cdot \left( {2 + 3 + ... + 100 + 101} \right)\]

        \( = 103 \cdot \frac{{\left( {2 + 101} \right) \cdot 100}}{2}\)

         \( = 103 \cdot \frac{{103 \cdot 100}}{2} = 530\,\,450\).