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\(E=-\dfrac{1}{3}\cdot\left(1+2+3\right)-\dfrac{1}{4}\left(1+2+3+4\right)-...-\dfrac{1}{50}\left(1+2+3+...+50\right)\)
\(=\dfrac{-1}{3}\cdot\dfrac{3\cdot4}{2}-\dfrac{1}{4}\cdot\dfrac{4\cdot5}{2}-...-\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)
\(=\dfrac{-4}{2}-\dfrac{5}{2}-...-\dfrac{51}{2}\)
\(=\dfrac{-\left(4+5+...+51\right)}{2}\)
\(=\dfrac{-\left(51+4\right)\cdot\dfrac{48}{2}}{2}=-\dfrac{1320}{2}=-660\)
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bạn ghi rất kho hiểu nên mình khuyên bạn ghi lại trên word rồi cop vào đây
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=> \(A=\frac{\left(\frac{49}{1}+\frac{48}{2}+...+\frac{1}{49}\right)}{50}=\frac{49}{50.1}+\frac{48}{50.2}+...+\frac{1}{50.49}\)
=> \(A=\frac{50-1}{50.1}+\frac{50-2}{50.2}+...+\frac{50-49}{50.49}\)
=> \(A=\left(\frac{50}{50.1}+\frac{50}{50.2}+...+\frac{50}{50.49}\right)-\left(\frac{1}{50.1}+\frac{2}{50.2}+...+\frac{49}{50.49}\right)\)
=> \(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\) ( có 49 số 1/50 )
=> \(A=1+\frac{1}{2}+...+\frac{1}{49}-\frac{49}{50}=\left(1-\frac{49}{50}\right)+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\)
=> \(A=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\)
Vậy A không phải là số tự nhiên
`Answer:`
\(C=-\frac{1}{3}.\left(1+2+3\right)-\frac{1}{4}.\left(1+2+3+4\right)-...-\frac{1}{50}.\left(1+2+3+...+50\right)\)
\(\Rightarrow C=-[\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{50}.\left(1+2+3+...+50\right)]\)
Ta có:
\(\frac{3.4}{2}=1+2+3\)
\(\frac{4.5}{2}=1+2+3+4\)
...
\(\frac{50.51}{2}=1+2+3+...+50\)
\(\Rightarrow C=-[\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{50}.\frac{50.51}{2}]\)
\(\Rightarrow C=-\left(\frac{4}{2}+\frac{5}{2}+...+\frac{51}{2}\right)\)
\(\Rightarrow C=-\frac{1}{2}.\left(4+5+...+51\right)\)
Đặt \(D=4+5+...+51\)
\(=\left(51+4\right).[\left(51-4\right):1+1]:2\)
\(=55.48:2\)
\(=1320\)
\(\Rightarrow C=-\frac{1}{2}.1320\)
\(\Rightarrow C=-660\)