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Nhận xét: 1/1.2.3 - 1/2.3.4 = 3/1.2.3.4, 1/2.3.4 - 1/3.4.5 =3/2.3.4.5,...,1/27.28.29 - 1/28.29.30
Gọi biểu thức phải tính bằng A,ta tính được:
3A=1/2.3 - 1/28.29.30 = 4059/28.29.30
vậy A = 1353/8120
\(c=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(c=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)\(=\)\(\dfrac{1}{6}-\dfrac{1}{24360}\)
\(C=\) \(\dfrac{4059}{24360}\)
Ta có:
\(C=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1353}{8120}\)
\(\Rightarrow C=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)
Vậy \(C=\dfrac{451}{8120}\)
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 \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}=\frac{451}{8120}\)
2P=\(\dfrac{2}{2}+\dfrac{2}{2^2}+...+\dfrac{2}{2^{100}}\)
2P=\(1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)
2P-P=\(\dfrac{1}{2}-\dfrac{1}{2^{100}}\)
P=\(\dfrac{1}{2}-\dfrac{1}{2^{100}}\)
\(P=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2P=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)\(\)
\(2P-P=1-\dfrac{1}{2^{100}}\)
\(P=\dfrac{2^{100}}{2^{100}}-\dfrac{1}{2^{100}}\)
\(P=\dfrac{2^{100}-1}{2^{100}}\)
P=1/1.2.3.4 +1/2.3.4.5 +1/3.4.5.6 +...+1/97.98.99.100
3P=3/1.2.3.4 +3/2.3.4.5 +3/3.4.5.6 +...+3/97.98.99.100
3P=1/1.2.3-1/2.3.4+1/2.3.4-1/3.4.5+................+1/97.98.99-1/98.99.100
3P = 1/1.2.3 - 1/98.99.100
3P =( 98.99.100-1.2.3)/1.2.3.98.99.100
P=( 98.99.100-1.2.3)/1.2.3.98.99.100.3
P=(98.33.50-1)/98.99.100.3
P= 161699/2910600
\(A=\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
\(A=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\right)\)
\(A=\left(1-\dfrac{1}{100}\right)-\left(\dfrac{\dfrac{1}{1.2}-\dfrac{1}{100.101}}{2}\right)\)
Bấm máy nha
\(B=\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+\dfrac{5}{3.4.5.6}+...+\dfrac{5}{98.99.100.101}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+...+\dfrac{101-98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4}{1.2.3.4}-\dfrac{1}{1.2.3.4}+\dfrac{5}{2.3.4.5}-\dfrac{2}{2.3.4.5}+...+\dfrac{101}{98.99.100.101}-\dfrac{98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{99.100.101}\right)\)
\(B=\dfrac{5}{3}.\dfrac{166649}{999900}\approx0,3\)
Đặt \(A=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
Ta có:
\(3A=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3A=\dfrac{1}{6}-\dfrac{1}{24360}\)
\(\Rightarrow3A=\dfrac{1353}{8120}\)
\(\Rightarrow A=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)
Vậy \(A=\dfrac{451}{8120}\)
Ta có: \(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)