Chứng tỏ rằng: A= 1/1.2.3 + 1/2.3.4 +1/3.4.5+….+1/20.21.22 > 57/231 20/10/2021 Bởi Reese Chứng tỏ rằng: A= 1/1.2.3 + 1/2.3.4 +1/3.4.5+….+1/20.21.22 > 57/231
Ta có : $A = \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+….+\dfrac{1}{20.21.22}$ $\to 2A = \dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+….+\dfrac{2}{20.21.22}$ $ = \dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+….+\dfrac{1}{20.21}-\dfrac{1}{21.22}$ $ = \dfrac{1}{1.2}-\dfrac{1}{21.22}$ $ = \dfrac{1}{2} – \dfrac{1}{462} = \dfrac{115}{231}$ $\to A = \dfrac{115}{462} > \dfrac{114}{462} = \dfrac{57}{231}$ Bình luận
Giải thích các bước giải: $A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{20.21.22}\\\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3}\\\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{2}{2.3.4}\\\dfrac{1}{3.4}-\dfrac{1}{4.5}=\dfrac{2}{3.4.5}\\…\\\dfrac{1}{20.21}-\dfrac{1}{21.22}=\dfrac{2}{20.21.22}\\\Rightarrow 2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+…+\dfrac{2}{20.21.22}\\=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+…+\dfrac{1}{20.21}-\dfrac{1}{21.22}\\=\dfrac{1}{2}-\dfrac{1}{21.22}\\=\dfrac{231}{462}-\dfrac{1}{462}=\dfrac{230}{462}=\dfrac{115}{231}\\\Rightarrow A=\dfrac{115}{231}.\dfrac{1}{2}=\dfrac{115}{462}\\\dfrac{57}{231}=\dfrac{114}{462}$Vì $114<115\Rightarrow \dfrac{114}{462}<\dfrac{115}{462}$Vậy $A>\dfrac{57}{231}$ Bình luận
Ta có : $A = \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+….+\dfrac{1}{20.21.22}$
$\to 2A = \dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+….+\dfrac{2}{20.21.22}$
$ = \dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+….+\dfrac{1}{20.21}-\dfrac{1}{21.22}$
$ = \dfrac{1}{1.2}-\dfrac{1}{21.22}$
$ = \dfrac{1}{2} – \dfrac{1}{462} = \dfrac{115}{231}$
$\to A = \dfrac{115}{462} > \dfrac{114}{462} = \dfrac{57}{231}$
Giải thích các bước giải:
$A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{20.21.22}\\
\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3}\\
\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{2}{2.3.4}\\
\dfrac{1}{3.4}-\dfrac{1}{4.5}=\dfrac{2}{3.4.5}\\
…\\
\dfrac{1}{20.21}-\dfrac{1}{21.22}=\dfrac{2}{20.21.22}\\
\Rightarrow 2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+…+\dfrac{2}{20.21.22}\\
=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+…+\dfrac{1}{20.21}-\dfrac{1}{21.22}\\
=\dfrac{1}{2}-\dfrac{1}{21.22}\\
=\dfrac{231}{462}-\dfrac{1}{462}=\dfrac{230}{462}=\dfrac{115}{231}\\
\Rightarrow A=\dfrac{115}{231}.\dfrac{1}{2}=\dfrac{115}{462}\\
\dfrac{57}{231}=\dfrac{114}{462}$
Vì $114<115\Rightarrow \dfrac{114}{462}<\dfrac{115}{462}$
Vậy $A>\dfrac{57}{231}$