Đáp án:
\[328350\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = {1^2} + {2^2} + {3^2} + {4^2} + ... + {98^2} + {99^2}\\
= 1.1 + 2.2 + 3.3 + 4.4 + .... + 98.98 + 99.99\\
= 1.\left( {2 - 1} \right) + 2.\left( {3 - 1} \right) + 3.\left( {4 - 1} \right) + 4\left( {5 - 1} \right) + .... + 98.\left( {99 - 1} \right) + 99.\left( {100 - 1} \right)\\
= \left( {1.2 + 2.3 + 3.4 + 4.5 + ..... + 98.99 + 99.100} \right) - \left( {1 + 2 + 3 + 4 + .... + 98 + 99} \right)\\
B = 1.2 + 2.3 + 3.4 + 4.5 + .... + 98.99 + 99.100\\
\Rightarrow 3B = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + .... + 98.99.3 + 99.100.3\\
\Leftrightarrow 3B = 1.2.3 + 2.3.\left( {4 - 1} \right) + 3.4.\left( {5 - 2} \right) + 4.5.\left( {6 - 3} \right) + .... + 98.99\left( {100 - 97} \right) + 99.100.\left( {101 - 98} \right)\\
\Leftrightarrow 3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100\\
\Leftrightarrow 3B = 99.100.101\\
\Leftrightarrow B = 33.100.101\\
\Rightarrow A = B - \left( {1 + 2 + 3 + .... + 99} \right)\\
= 33.100.101 - \frac{{\left( {1 + 99} \right).99}}{2}\\
= 33.100.101 - \frac{{100.99}}{2}\\
= 33.100\left( {101 - \frac{3}{2}} \right)\\
= 328350
\end{array}\)
