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TL
10 tháng 5 2017
hôm qua cô giảng cho mình bài này không cần tính đâu
Gọi tổng là A
A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{17.18.19}\)
2A=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{17.18.19}\)
2A=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{17.18}-\dfrac{1}{18.19}\)
2A=\(\dfrac{1}{2}-\dfrac{1}{18.19}\)
A=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{18.19}\right)\)
A=\(\dfrac{1}{2}.\dfrac{18.19-2}{2.18.19}\) < \(\dfrac{1}{4}\)
A=\(\dfrac{18.19-2}{2.2.18.19}\) < \(\dfrac{18.19}{2.2.18.19}\)
\(\Rightarrow\) A<\(\dfrac{1}{4}\)
9 tháng 5 2017
\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)<\(\dfrac{1}{4}\)
Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)
2.A=2.(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\))
2. A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+\(\dfrac{2}{3.4.5}\)+...+\(\dfrac{2}{17.18.19}\)
2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+ ...+\(\dfrac{1}{17.18}\)-\(\dfrac{1}{18.19}\)
2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{18.19}\)=\(\dfrac{85}{171}\)
A=\(\dfrac{85}{171}\):2=\(\dfrac{85}{342}\)
Ta cũng có: \(\dfrac{1}{4}\) = \(\dfrac{171}{684}\); \(\dfrac{85}{342}\) = \(\dfrac{170}{684}\)
Vì 170 < 171 ( \(\dfrac{170}{684}\) < \(\dfrac{171}{684}\) )
Vậy \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{17.18.19}\) < \(\dfrac{1}{4}\)
NT
1
8 tháng 4 2023
\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{17\cdot18\cdot19}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{17\cdot18}-\dfrac{1}{18\cdot19}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{342}\right)=\dfrac{1}{2}\cdot\dfrac{85}{171}=\dfrac{85}{342}\)
24 tháng 9 2021
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
TA
0
30 tháng 6 2018
Đặt A = 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4 . 5 + ... + 17 . 18 . 19 + 18 . 19 . 20
=> 4A = 1 . 2 . 3 . 4 + 2 . 3 . 4 . 4 + 3 . 4 . 5 . 4 + ... + 17 . 18 . 19 . 4 + 18 . 19 . 20 . 4
4A = 1 . 2 . 3 . 4 + 2 . 3 . 4 . ( 5 - 1 ) + 3 . 4 . 5 . ( 6 - 2 ) + ... + 17 . 18 . 19 . ( 20 - 16 ) + 18 . 19 . 20 . ( 21 - 17 )
4A = 1 . 2 . 3 . 4 + 2 . 3 . 4 . 5 - 1 . 2 . 3 . 4 + 3 . 4 . 5 . 6 - 2 . 3 . 4 . 5+ ... + 17.18.19.20 - 16.17.18.19 + 18.19.20.21 -17.18.19.20
4A = 18 . 19 . 20 . 21
=> A = 18 . 19 . 20 . 21 : 4
A = 35 910
30 tháng 6 2018
Đặt M = 1.2.3+2.3.4 + 3.4.5+...+17.18.19+18.19.20
=> 4M = 1.2.3.4+2.3.4.4+3.4.5.4+...+17.18.19.4+18.19.20
4M = 1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+...+17.18.19.(20-16)+18.19.20.(21-17)
4M = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ...+17.18.19.20 - 16.17.18.19 + 18.19.20.21 - 17.18.19.20
4M = ( 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + ...+ 17.18.19.20+18.19.20.21) - (1.2.3.4+2.3.4.5+...+16.17.18.19+17.18.19.20)
4M = 18.19.20.21
\(M=\frac{18.19.20.21}{4}\)
MM
6
L
9 tháng 4 2018
A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19
=> 4A = 4(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
=> 4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
=> 4A = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
=> 4A = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
=> 4A = 17.18.19.20
=> 4A = 116280
=> A = 29070
Vậy A = 29070
Mà cxh là gì z
9 tháng 4 2018
A = 1.2.3 + 2.3.4 + ... + 17.18.19
4A = 1.2.3.4 + 2.3.4.4 + ... + 17.18.19.4
= 1.2.3.4 + 2.3.4.(5-1) + ... + 17.18.19 (20 - 16) = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 +...+ 17.18.19.20 - 16.17.18.19
4A = 17.18.19.20
A= (17.18.19.20) : 4 = 29070
25 tháng 12 2021
Ta có \(k\left(k+1\right)\left(k+2\right)=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\cdot4\)
\(=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left[\left(k+3\right)-\left(k-1\right)\right]\\ =\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\dfrac{1}{4}\left(k-1\right)k\left(k+1\right)\left(k+2\right)\)
Từ đó ta được \(S=\dfrac{1}{4}\cdot1\cdot2\cdot3\cdot4-\dfrac{1}{4}\cdot0\cdot1\cdot2\cdot3+...+\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12-\dfrac{1}{4}\cdot8\cdot9\cdot10\cdot11\\ \Leftrightarrow S=\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12\\ \Leftrightarrow4S+1=9\cdot10\cdot11\cdot12+1=11881=109^2\left(đpcm\right)\)
KK
1
YA
18 tháng 11 2016
Ta có : \(k\left(k+1\right)\left(k+2\right)=\frac{1}{4}k\left(k+1\right)\left(k+2\right).4\)
\(=\frac{1}{4}k\left(k+1\right)\left(k+2\right)\left[\left(k+3\right)-\left(k-1\right)\right]\)
\(=\frac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\frac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k-1\right)\)
=> 4S = 1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+k(k+1)(k+2)(k+3)-k(k+1)(k+2)(k-1)
\(=k\left(k+1\right)\left(k+2\right)\left(k+3\right)\)
=> \(4S+1=k\left(k+1\right)\left(k+2\right)\left(k+3\right)+1\)
\(=\left[k\left(k+3\right)\right]\left[\left(k+1\right)\left(k+2\right)\right]+1\)
\(=\left[\left(k^2+3k\right)\left(k^2+k+2k+2\right)\right]+1\)
Đặt \(t=k^2+3k\)
\(=>4S+1=t\left(t+2\right)+1\)
= \(t^2+2t+1\)
\(=\left(t+1\right)^2\)
\(=>4S+1=\left(k^2=3k\right)^2=>4S+1\) là số chính phương
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{17.18.19}\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{17.18}-\frac{1}{18.19}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{18.19}\right)=\frac{1}{4}-\frac{1}{2.18.19}< \frac{1}{4}\)
oh help