Tính nhanh : A = 1 mũ 2 + 2 mũ 2 + 3 mũ 2 + …+98 mũ 2+ 99 mũ 2

Đáp án:

\[328350\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
A = {1^2} + {2^2} + {3^2} + {4^2} + … + {98^2} + {99^2}\\
 = 1.1 + 2.2 + 3.3 + 4.4 + …. + 98.98 + 99.99\\
 = 1.\left( {2 – 1} \right) + 2.\left( {3 – 1} \right) + 3.\left( {4 – 1} \right) + 4\left( {5 – 1} \right) + …. + 98.\left( {99 – 1} \right) + 99.\left( {100 – 1} \right)\\
 = \left( {1.2 + 2.3 + 3.4 + 4.5 + ….. + 98.99 + 99.100} \right) – \left( {1 + 2 + 3 + 4 + …. + 98 + 99} \right)\\
B = 1.2 + 2.3 + 3.4 + 4.5 + …. + 98.99 + 99.100\\
 \Rightarrow 3B = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + …. + 98.99.3 + 99.100.3\\
 \Leftrightarrow 3B = 1.2.3 + 2.3.\left( {4 – 1} \right) + 3.4.\left( {5 – 2} \right) + 4.5.\left( {6 – 3} \right) + …. + 98.99\left( {100 – 97} \right) + 99.100.\left( {101 – 98} \right)\\
 \Leftrightarrow 3B = 1.2.3 + 2.3.4 – 1.2.3 + 3.4.5 – 2.3.4 + …. + 98.99.100 – 97.98.99 + 99.100.101 – 98.99.100\\
 \Leftrightarrow 3B = 99.100.101\\
 \Leftrightarrow B = 33.100.101\\
 \Rightarrow A = B – \left( {1 + 2 + 3 + …. + 99} \right)\\
 = 33.100.101 – \frac{{\left( {1 + 99} \right).99}}{2}\\
 = 33.100.101 – \frac{{100.99}}{2}\\
 = 33.100\left( {101 – \frac{3}{2}} \right)\\
 = 328350
\end{array}\)