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a) \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{27.28.29}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{27.28}-\frac{1}{28.29}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{28.29}\right)\)
\(=\frac{1}{2}.\frac{405}{812}=\frac{405}{1624}\)
Vậy giá trị của biểu thức \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{27.28.29}=\frac{405}{1624}\)
b) \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+....+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
\(=\frac{1}{3}\cdot\frac{1353}{8120}=\frac{451}{8120}\)
Vậy giá trị của biểu thức \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}=\frac{451}{8120}\)
đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{189}{760}\)
Đặt \(B=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+...+\frac{3}{19}-\frac{3}{20}\)
\(=3-\frac{3}{20}=\frac{57}{20}\)
\(D=A-B=\frac{189}{760}-\frac{57}{20}=-\frac{1977}{760}\)
Gọi \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)là A
\(\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)là B
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{20}\right)\right]\)
\(A=\frac{1}{2}.\frac{19}{20}\)
\(A=\frac{19}{40}\)
\(B=\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)
\(B=\left(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}\right)\)
\(B=\left[3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)\right]\)
\(B=\left[3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(B=\left[3.\left(\frac{19}{20}\right)\right]\)
\(B=\frac{57}{20}\)
Vậy A - B = \(\frac{19}{40}-\frac{57}{20}\)
\(=-\frac{95}{40}=-\frac{19}{8}\)
Nếu đúng thì k nha
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{17.18.19}\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{17.18}-\frac{1}{18.19}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{18.19}\right)=\frac{1}{4}-\frac{1}{2.18.19}< \frac{1}{4}\)
tao có:
2p=2/1.2.3+2/2.3.4+...+2/n.n(+1)n(n+2)
2p=3-1/1.2.3+4-2/1.2.3+...+(n+2)-n/n.(n+1).(n+2)
2p=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+(n+2)/n.(n+1).(n+2)-n/n.(n+1).(n+2)
2p=1/1.2-1/2.3+1/2.3-1/3.4+...+1/n.(n+1)-1/(n+1).(n+2)
2p=1/1.2-1/(n+1).(n+2)
2p=(n+!).(n+2)-2/(2n+2).(n+2)
suy ra p=(n+1).(n+2)-2/(2n+2).(2n+4)
2s=3-1/1.2.3+4-2/1.2.3+...+50-48/48.49.50
2s=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+50/49.50.48-48/48.50.49
2s=1/1.2-1/2.3+1/2.3-1/3.4+...+1/48.49-1/49.50
2s=1/1.2-1/49.50
'2s=1/2-1/2450
2s=1225/2450-1/2450
2s=1224/2450
s=612/1225
\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)1
\(P=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)}{2}\)
S cx tinh giong v
gọi A=................................
=>2A=2/1.2.3+2/2.3.4+.....+2/8.9.10
2A=1/1.2-1/2.3+1/2.3-...+1/8.9-1/9.10
2A=1/1.2-1/9.10=22/45 =>A=11/45
\(=\dfrac{1}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\right)\)
\(=\dfrac{1353}{8120}\)