Toán Lớp 6: Bài 2: Cho $\textit{M}$ = $\dfrac{1}{1. 2}$ + $\dfrac{1}{1. 2. 3}$ + $\dfrac{1}{1. 2. 3. 4}$ + ….. + $\dfrac{1}{1. 2. 3….100}$. Chứng minh rằng $\textit{M}$ < 1
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Toán Lớp 6: Bài 2: Cho $\textit{M}$ = $\dfrac{1}{1. 2}$ + $\dfrac{1}{1. 2. 3}$ + $\dfrac{1}{1. 2. 3. 4}$ + ….. + $\dfrac{1}{1. 2. 3….100}$. Chứng minh rằng $\textit{M}$ < 1
Comments ( 1 )
Giải đáp + Lời giải và giải thích chi tiết:
Bài 2$\text{:}$
Ta có$\text{: $\dfrac{1}{1. 2. 3}$ = $\dfrac{1}{2. 3}$}$
$\text{$\dfrac{1}{1. 2. 3. 4}$ < $\dfrac{1}{3. 4}$}$
$\text{$\dfrac{1}{1. 2. 3. 4. 5}$ < $\dfrac{1}{4. 5}$}$
$\text{….}$
$\text{$\dfrac{1}{1. 2. 3…100}$ < $\dfrac{1}{99. 100}$}$
⇒ $\textit{M}$ $\text{< $\dfrac{1}{1. 2}$ + $\dfrac{1}{2. 3}$ + $\dfrac{1}{3. 4}$ + …. + $\dfrac{1}{99. 100}$}$
$\text{= $\dfrac{1}{1}$ – $\dfrac{1}{2}$ + $\dfrac{1}{2}$ – $\dfrac{1}{3}$ + $\dfrac{1}{3}$ – $\dfrac{1}{4}$ + …. + $\dfrac{1}{99}$ – $\dfrac{1}{100}$}$= 1 – $\dfrac{1}{100}$ < 1
$\text{⇒}$ $\textit{M}$ $\text{< 1}$
Vậy ta có điều phải chứng minh.