Tính nhanh : A = 1 mũ 2 + 2 mũ 2 + 3 mũ 2 + …+98 mũ 2+ 99 mũ 2 19/07/2021 Bởi Elliana 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: \(\begin{array}{l}A = {1^2} + {2^2} + {3^2} + {4^2} + … + {98^2} + {99^2}\\ = 1.1 + 2.2 + 3.3 + 4.4 + …. + 98.98 + 99.99\\ = 1.\left( {2 – 1} \right) + 2.\left( {3 – 1} \right) + 3.\left( {4 – 1} \right) + 4\left( {5 – 1} \right) + …. + 98.\left( {99 – 1} \right) + 99.\left( {100 – 1} \right)\\ = \left( {1.2 + 2.3 + 3.4 + 4.5 + ….. + 98.99 + 99.100} \right) – \left( {1 + 2 + 3 + 4 + …. + 98 + 99} \right)\\B = 1.2 + 2.3 + 3.4 + 4.5 + …. + 98.99 + 99.100\\ \Rightarrow 3B = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + …. + 98.99.3 + 99.100.3\\ \Leftrightarrow 3B = 1.2.3 + 2.3.\left( {4 – 1} \right) + 3.4.\left( {5 – 2} \right) + 4.5.\left( {6 – 3} \right) + …. + 98.99\left( {100 – 97} \right) + 99.100.\left( {101 – 98} \right)\\ \Leftrightarrow 3B = 1.2.3 + 2.3.4 – 1.2.3 + 3.4.5 – 2.3.4 + …. + 98.99.100 – 97.98.99 + 99.100.101 – 98.99.100\\ \Leftrightarrow 3B = 99.100.101\\ \Leftrightarrow B = 33.100.101\\ \Rightarrow A = B – \left( {1 + 2 + 3 + …. + 99} \right)\\ = 33.100.101 – \frac{{\left( {1 + 99} \right).99}}{2}\\ = 33.100.101 – \frac{{100.99}}{2}\\ = 33.100\left( {101 – \frac{3}{2}} \right)\\ = 328350\end{array}\) Bình luận
Đáp án:
\[328350\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = {1^2} + {2^2} + {3^2} + {4^2} + … + {98^2} + {99^2}\\
= 1.1 + 2.2 + 3.3 + 4.4 + …. + 98.98 + 99.99\\
= 1.\left( {2 – 1} \right) + 2.\left( {3 – 1} \right) + 3.\left( {4 – 1} \right) + 4\left( {5 – 1} \right) + …. + 98.\left( {99 – 1} \right) + 99.\left( {100 – 1} \right)\\
= \left( {1.2 + 2.3 + 3.4 + 4.5 + ….. + 98.99 + 99.100} \right) – \left( {1 + 2 + 3 + 4 + …. + 98 + 99} \right)\\
B = 1.2 + 2.3 + 3.4 + 4.5 + …. + 98.99 + 99.100\\
\Rightarrow 3B = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + …. + 98.99.3 + 99.100.3\\
\Leftrightarrow 3B = 1.2.3 + 2.3.\left( {4 – 1} \right) + 3.4.\left( {5 – 2} \right) + 4.5.\left( {6 – 3} \right) + …. + 98.99\left( {100 – 97} \right) + 99.100.\left( {101 – 98} \right)\\
\Leftrightarrow 3B = 1.2.3 + 2.3.4 – 1.2.3 + 3.4.5 – 2.3.4 + …. + 98.99.100 – 97.98.99 + 99.100.101 – 98.99.100\\
\Leftrightarrow 3B = 99.100.101\\
\Leftrightarrow B = 33.100.101\\
\Rightarrow A = B – \left( {1 + 2 + 3 + …. + 99} \right)\\
= 33.100.101 – \frac{{\left( {1 + 99} \right).99}}{2}\\
= 33.100.101 – \frac{{100.99}}{2}\\
= 33.100\left( {101 – \frac{3}{2}} \right)\\
= 328350
\end{array}\)