Cho A=1.2.3+2.3.4+...+111.112.113. Vậy A=?
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TN
8 tháng 3 2016
<=>4S=1.2.3.4 + 2.3.4.4+3.4.5.4+.....+8.9.10.4
<=>4S =1.2.3.4 + 2.3.4.(5-1) + 3.4.5.(6-2)+......+8.9.10.(11-7)
<=>4S=1.2.3.4 + 2.3.4.5 -1.2.3.4+3.4.5.6- 2.3.4.5+......+8.9.10.11 - 7.8.9.10
<=> 4S=8.9.10.11
<=>S=1980
KV
19 tháng 8 2023
Sửa đề: \(A=1.2.3+2.3.4+3.4.5+...+8.9.10\)
\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+...+8.9.10.4\)
\(=2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+8.9.10.\left(11-7\right)\)
\(=2.3.4+2.3.4.5-2.3.4+3.4.5.6-2.3.4.5+...+8.9.10.11-7.8.9.10\)
\(=8.9.10.11\)
\(=7920\)
\(\Rightarrow A=\dfrac{7920}{4}=1980\)
19 tháng 8 2023
Có: A=1.2.3+2.3.4+4.5.6+...+8.9.10
4A=1.2.3.4 + 2.3.4.4+...+8.9.10.4
4A=1.2.3.(4-0)+2.3.4.(5-1)+...+8.9.10.(11-7)
4A=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+...+8.9.10.11−7.8.9.1
4A=(1.2.3.4+2.3.4.5+...+8.9.10.11)−(0.1.2.3+1.2.3.4+...+7.8.9.10)
BJ
1
9 tháng 5 2018
\(A=\frac{4}{1.2.3}+\frac{4}{2.3.4}+\frac{4}{3.4.5}+...+\frac{4}{24.25.26}\)
\(=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{24.25}-\frac{1}{25.26}\right)\) \(=2.\frac{2}{1.2.3}+2.\frac{2}{2.3.4}+2.\frac{2}{3.4.5}+...+2.\frac{2}{24.25.26}\)
\(=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{24.25}-\frac{1}{25.26}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{25.26}\right)=2.\frac{162}{325}=\frac{324}{325}\)
NV
0
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 \(A< \dfrac{1}{4}\)
T
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 1 2023
A= 1.2.3 + 2.3.4 + 3.4.5 +.....+ 98.99.100
4A = 98.99.100.4 + .....+ 3.4.5.4 + 2.3.4.4 + 1.2.3.4
4A = 98.99.100.(101-97) +... + 2.3.4.(5-1) + 1.2.3.4
4A = 98.99.100.101 - 97.98.99.100+......+2.3.4.5 - 1.2.3.4 + 1.2.3.4
4A = 98.99.100.101
A = 98.99.100.101 : 4
A = 24497550
NT
1
NN
3
NT
17 tháng 5 2017
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{2014.2015.2016}\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+.....+\left(\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2015.2016}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.2015.2016}< \dfrac{1}{4}\\ \Rightarrow A< \dfrac{1}{4}\)
17 tháng 5 2017
Giải:
Ta có: \(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2014.2015.2016}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2014.2015.2016}\right)\)
\(=\dfrac{1}{2}\)\(\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2015.2016}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{1.2}-\dfrac{1}{2}.\dfrac{1}{2015.2016}=\dfrac{1}{4}-\) \(\dfrac{1}{2.2015.2016}\)
Mà \(\dfrac{1}{2.2015.2016}>0\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{2.2015.2016}< \dfrac{1}{4}\)
Vậy \(A< \dfrac{1}{4}\)
1 số siêu khủng: Math ERROR