A=$\frac{m-1}{1}$ + $\frac{m-2}{2}$ +...+ $\frac{2}{m-2}$ + $\frac{1}{m-1}$ ; B=$\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ +...+ $\frac{1}{n}$. Tính $\frac{A}{B}$
1 câu trả lời
`#huy`
Ta có:
`A=(n/1+n/2+...+n/(n-2)+n/(n-1))-(\underbrace{1+1+...+1}_{n-1})`
`=n(1/1+1/2+1/(n-2)+1/(n-1))-(n-1)`
`=n(1/2+...+1/(n-2)+1/(n-1))+1`
`=n(1/2+...+1/(n-2)+1/(n-1)+1/n)`
`=nB`
`=>A/B=n`
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