Tính nhanh giá trị các biểu thức sau: f) F = 1^ 2 + 2^ 2 + 3^ 2 + . . . + 99 ^2 + 100 ^2 .

f) \(F = {1^2} + {2^2} + {3^2} + ... + {99^2} + {100^2}\)

\( = 1 \cdot \left( {2 - 1} \right) + 2 \cdot \left( {3 - 1} \right) + 3 \cdot \left( {4 - 1} \right) + ... + 99 \cdot \left( {100 - 1} \right) + 100 \cdot \left( {101 - 1} \right)\)

\[ = 1 \cdot 2 - 1 \cdot 1 + 2 \cdot 3 - 2 \cdot 1 + 3 \cdot 4 - 3 \cdot 1 + ... + 99 \cdot 100 - 99 \cdot 1 + 100 \cdot 101 - 100 \cdot 1\]

\[ = 1 \cdot 2 - 1 + 2 \cdot 3 - 2 + 3 \cdot 4 - 3 + ... + 99 \cdot 100 - 99 + 100 \cdot 101 - 100\]

\[ = \left( {1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 - 3 + ... + 99 \cdot 100 + 100 \cdot 101} \right) - \left( {1 + 2 + ... + 99 + 100} \right)\]

\[ = \frac{{100 \cdot 101 \cdot 102}}{3} - \frac{{100 \cdot \left( {100 + 1} \right)}}{2}\]                   (tương tự câu d)

\[ = \frac{{100 \cdot 101 \cdot 102}}{3} - \frac{{100 \cdot 101}}{2}\]

\[ = \frac{{100 \cdot 101 \cdot 102 \cdot 2 - 100 \cdot 101 \cdot 3}}{6}\]

\( = \frac{{100 \cdot 101 \cdot 201}}{6} = 338\,\,350.\)