chưng tỏ 1/1.2.3 +1/2.3.4 + 1/3.4.5+...+1/98.99.100=4949/19800
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DT
1
IA
4 tháng 3 2020
\(\frac{2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\right)}{2}\)
(\(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\)) : 2
(\(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)) : 2
4 tháng 3 2017
a) Ta có: \(3xy+x-3y=6\)
\(\Rightarrow x\left(3y+1\right)-3y=6\)
\(\Rightarrow x\left(3y+1\right)-\left(3y+1\right)=5\)
\(\Rightarrow\left(x-1\right)\left(3y+1\right)=5\)
Ta có bảng sau:
....
b) Ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
\(\Rightarrow\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}=\frac{4949}{19800}\left(đpcm\right)\)
Vậy...
KH
12 tháng 5 2017
* Chứng tỏ
Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
= \(\dfrac{1}{1.2.3}.\dfrac{2}{2}+\dfrac{1}{2.3.4}.\dfrac{2}{2}+...+\dfrac{1}{98.99.100}.\dfrac{2}{2}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)
= \(\dfrac{4849}{19800}\)
HT
2 tháng 8 2015
1/1.2.3 + 1/2.3.4 +....+1/98.99.100
= 1/2 . (3-1/1.2.3 + 4-2/2.3.4 +....+ 100-98/98.99.100)
= 1/2 . (3/1.2.3 -1/1.2.3 + 4/2.3.4 - 2/2.3.4 +.......+ 100/98.99.100 - 98/98.99.100)
= 1/2 . (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +......+ 1/98.99 - 1/99.100)
= 1/2 . (1/2 - 1/9900)
= 1/2 . 4949/9900
= 4949/19800
Bạn cho sai đề rồi !
Sửa : Chứng tỏ : \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{4949}{9900}\)
Ta có : \(VT=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(=\frac{99.100-2}{2.99.100}\)
\(=\frac{4949}{9900}=VP\)
Study well ! >_<