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\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{17\cdot18\cdot19}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{17\cdot18}-\dfrac{1}{18\cdot19}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{342}\right)=\dfrac{1}{2}\cdot\dfrac{85}{171}=\dfrac{85}{342}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{17.18.19}\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{17.18}-\frac{1}{18.19}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{18.19}\right)=\frac{1}{4}-\frac{1}{2.18.19}< \frac{1}{4}\)
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
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
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<\(\dfrac{1}{4}\)
\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)<\(\dfrac{1}{4}\)
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 \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{17.18.19}\) < \(\dfrac{1}{4}\)
\(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}
B=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}< 3\)
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
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
\(M=\frac{18.19.20.21}{4}\)