CMR : A = 1/1.2.3 + 1/2.3.4 + ... + 1/18.19.20 < 1/4
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NL
2
AM
11 tháng 7 2015
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{1}{4}-\frac{1}{2.19.20}
6 tháng 3 2018
B=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}< 3\)
MV
8 tháng 5 2017
Nhận thấy: \(\dfrac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\\ =\dfrac{2}{2\cdot n\cdot\left(n+1\right)\cdot\left(n+2\right)}\\ =\dfrac{2+n-n}{2n\cdot\left(n+1\right)\cdot\left(n+2\right)}\\ =\dfrac{1}{2}\cdot\left[\dfrac{2+n-n}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\right]\\ =\dfrac{1}{2}\cdot\left[\dfrac{2+n}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}-\dfrac{n}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\right]\\ =\dfrac{1}{2}\cdot\left[\dfrac{1}{n\cdot\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\cdot\left(n+2\right)}\right]\)
\(\Rightarrow A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{18\cdot19\cdot20}\\ =\dfrac{1}{2}\cdot\left[\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\right]\\ =\dfrac{1}{2}\cdot\left[\dfrac{1}{1\cdot2}-\dfrac{1}{19\cdot20}\right]\\ =\dfrac{1}{4}-\dfrac{1}{760}< \dfrac{1}{4}\)
Vậy \(A< \dfrac{1}{4}\)
DT
0
TA
0
CC
1
19 tháng 3 2016
a chứng minh được bài toán tổng quát sau
2/[(n-1)n(n+1)] = 1/[(n-1)n] - 1/[n(n+1)]
Áp dụng:
ta có 2A = 1/(1.2) - 1/ (2.3) +1/(2.3) - 1/(3.4) + ...+ 1/18.19 - 1/19.20
= 1/(1.2) - 1/(19.20) = [190 - 1] / 19.20 = 189/380
=> A = 189/ 760 < 1/4
HT
0
T
10 tháng 8 2016
=(1/1-1/2-1/3)+(1/2-1/3-1/4)+...+(1/18-1/19-1/20)
=1/1-1/20=19/20
k nha
TB
4
6 tháng 8 2019
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=\frac{1}{2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}< \frac{1}{2}\)
\(2S< \frac{1}{2}\)
\(\Rightarrow S< \frac{1}{4}\) (ĐPCM)
CC
6 tháng 8 2019
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{4}-\frac{1}{760}\)
=> S < \(\frac{1}{4}\)( vì 1/4 > 0)
A = 1/1.2.3 + 1/2.3.4 + ... + 1/18.19.20
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(A=\frac{1}{4}-\frac{1}{2.19.20}< \frac{1}{4}\)