CMR A= 1^3 + 2^3 + … + 100^3 chia hết cho B=1+2+3+…+100 25/07/2021 Bởi Maya CMR A= 1^3 + 2^3 + … + 100^3 chia hết cho B=1+2+3+…+100
Giải thích các bước giải: Ta có: $B = 1 + 2 + 3 + … + 100 = \frac{{\left( {1 + 100} \right).100}}{2} = 5050$ Để CM $A \vdots B \Leftrightarrow \left\{ \begin{array}{l}A \vdots 50\\A \vdots 101\end{array} \right.\left( {5050 = 50.101;\left( {50,101} \right) = 1} \right)$ $\begin{array}{l} + )CM:A \vdots 50\\A = {1^3} + {2^3} + {3^3} + … + {100^3}\\ \Leftrightarrow A = \left( {{1^3} + {{99}^3}} \right) + \left( {{2^3} + {{98}^3}} \right) + … + \left( {{{49}^3} + {{51}^3}} \right) + {50^3} + {100^3}\\ \Leftrightarrow A = \left( {1 + 99} \right)\left( {{1^2} – 1.99 + {{99}^2}} \right) + \left( {2 + 98} \right)\left( {{2^2} – 2.98 + {{98}^2}} \right) + … + \left( {49 + 51} \right)\left( {{{49}^2} – 49.51 + {{51}^2}} \right) + {50^3} + {100^3}\\ \Leftrightarrow A = 100\left( {{1^2} – 1.99 + {{99}^2}} \right) + 100\left( {{2^2} – 2.98 + {{98}^2}} \right) + … + 100\left( {{{49}^2} – 49.51 + {{51}^2}} \right) + {50^3} + {100^3}\\ \Leftrightarrow A = 100\left( {{1^2} – 1.99 + {{99}^2} + {2^2} – 2.98 + {{98}^2} + {{49}^2} – 49.51 + {{51}^2}} \right) + {50^3} + {100^3}\\ \Rightarrow A \vdots 50\\ + )CM:A \vdots 101\\A = {1^3} + {2^3} + {3^3} + … + {100^3}\\ \Leftrightarrow A = \left( {{1^3} + {{100}^3}} \right) + \left( {{2^3} + {{99}^3}} \right) + … + \left( {{{50}^3} + {{51}^3}} \right)\\ \Leftrightarrow A = \left( {1 + 100} \right)\left( {{1^2} – 1.100 + {{100}^2}} \right) + \left( {2 + 99} \right)\left( {{2^2} – 2.99 + {{99}^2}} \right) + … + \left( {50 + 51} \right)\left( {{{50}^2} – 50.51 + {{51}^2}} \right)\\ \Leftrightarrow A = 101\left( {{1^2} – 1.100 + {{100}^2}} \right) + 101\left( {{2^2} – 2.99 + {{99}^2}} \right) + … + 101\left( {{{50}^2} – 50.51 + {{51}^2}} \right)\\ \Leftrightarrow A = 101\left( {{1^2} – 1.100 + {{100}^2} + {2^2} – 2.99 + {{99}^2} + {{50}^2} – 50.51 + {{51}^2}} \right)\\ \Rightarrow A \vdots 101\end{array}$ Như vậy ta cm được $A \vdots 5050$ hay $A \vdots B$ Bình luận
Giải thích các bước giải:
Ta có:
$B = 1 + 2 + 3 + … + 100 = \frac{{\left( {1 + 100} \right).100}}{2} = 5050$
Để CM $A \vdots B \Leftrightarrow \left\{ \begin{array}{l}
A \vdots 50\\
A \vdots 101
\end{array} \right.\left( {5050 = 50.101;\left( {50,101} \right) = 1} \right)$
$\begin{array}{l}
+ )CM:A \vdots 50\\
A = {1^3} + {2^3} + {3^3} + … + {100^3}\\
\Leftrightarrow A = \left( {{1^3} + {{99}^3}} \right) + \left( {{2^3} + {{98}^3}} \right) + … + \left( {{{49}^3} + {{51}^3}} \right) + {50^3} + {100^3}\\
\Leftrightarrow A = \left( {1 + 99} \right)\left( {{1^2} – 1.99 + {{99}^2}} \right) + \left( {2 + 98} \right)\left( {{2^2} – 2.98 + {{98}^2}} \right) + … + \left( {49 + 51} \right)\left( {{{49}^2} – 49.51 + {{51}^2}} \right) + {50^3} + {100^3}\\
\Leftrightarrow A = 100\left( {{1^2} – 1.99 + {{99}^2}} \right) + 100\left( {{2^2} – 2.98 + {{98}^2}} \right) + … + 100\left( {{{49}^2} – 49.51 + {{51}^2}} \right) + {50^3} + {100^3}\\
\Leftrightarrow A = 100\left( {{1^2} – 1.99 + {{99}^2} + {2^2} – 2.98 + {{98}^2} + {{49}^2} – 49.51 + {{51}^2}} \right) + {50^3} + {100^3}\\
\Rightarrow A \vdots 50\\
+ )CM:A \vdots 101\\
A = {1^3} + {2^3} + {3^3} + … + {100^3}\\
\Leftrightarrow A = \left( {{1^3} + {{100}^3}} \right) + \left( {{2^3} + {{99}^3}} \right) + … + \left( {{{50}^3} + {{51}^3}} \right)\\
\Leftrightarrow A = \left( {1 + 100} \right)\left( {{1^2} – 1.100 + {{100}^2}} \right) + \left( {2 + 99} \right)\left( {{2^2} – 2.99 + {{99}^2}} \right) + … + \left( {50 + 51} \right)\left( {{{50}^2} – 50.51 + {{51}^2}} \right)\\
\Leftrightarrow A = 101\left( {{1^2} – 1.100 + {{100}^2}} \right) + 101\left( {{2^2} – 2.99 + {{99}^2}} \right) + … + 101\left( {{{50}^2} – 50.51 + {{51}^2}} \right)\\
\Leftrightarrow A = 101\left( {{1^2} – 1.100 + {{100}^2} + {2^2} – 2.99 + {{99}^2} + {{50}^2} – 50.51 + {{51}^2}} \right)\\
\Rightarrow A \vdots 101
\end{array}$
Như vậy ta cm được $A \vdots 5050$ hay $A \vdots B$