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\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)
Tính:
S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+...+\frac{1}{98.99.100}\)
\(2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(2S=\frac{1}{2}-\frac{1}{9900}\)
\(2S=\frac{4949}{9900}\)
\(S=\frac{4949}{19800}\)
Ta xét : \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)
\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)
...
\(\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)
Ta có : 2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
=> 2S = \(\frac{1}{1.2}-\frac{1}{99.100}\)
=> 2S = \(\frac{4949}{9900}\)
=> S = \(\frac{4949}{19800}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
A=1/1.2.3+1/2.3.4+...+1/98.99.100
2A=2/1.2.3+2/2.3.4+...+2/98.99.100
2A=1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100
2A=1/1.2-1/99.100
2A=1/2-1/9900
2A=4949/9900
A=4949/19800
A=1/1.2.3+1/2.3.4+...+1/98.99.100
2A=2/1.2.3+2/2.3.4+...+2/98.99.100
2A=1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100
2A=1/1.2-1/99.100
2A=1/2-1/9900
2A=4949/9900
A=4949/19800
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}=\frac{4950}{9900}-\frac{1}{9900}=\frac{4949}{9900}\)
Ta có:
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(\Rightarrow\frac{1}{2}A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{98.99.100}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{98.99}-\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{99.100}\)
\(\Rightarrow A=\frac{4949}{9900}.\frac{1}{2}=\frac{4949}{19800}\)
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(A=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
chỗ nãy rồi bạn tự tính tiếp
A = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + ...+ \(\frac{1}{98.99.100}\)
2A = \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + ... + \(\frac{2}{98.99.100}\)
2A = \(\frac12\).(\(\frac{2}{1.3}\)) + \(\frac13\).(\(\frac{2}{2.4}\)) + ... + \(\frac{1}{99}\).(\(\frac{2}{98.100}\))
2A = \(\frac12\).(\(\frac11-\frac13\)) + \(\frac13\).(\(\frac12-\frac14\)) + ...+ \(\frac{1}{99}\).(\(\frac{1}{98}-\frac{1}{100}\))
2A = \(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + ...+\(\frac{1}{98.99}\) - \(\frac{1}{99.100}\)
2A = \(\frac12-\frac{1}{9900}\)
2A = \(\frac{4949}{9900}\)
A = \(\frac{4949}{9900}\) : 2
A = 4949/19800
Câu 2:
2Q = \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + ... + \(\frac{2}{98.99.100}\)
2Q = \(\frac12\).(\(\frac{2}{1.3}\)) + \(\frac13\).(\(\frac{2}{2.4}\)) + ... + \(\frac{1}{99}\).(\(\frac{2}{98.100}\))
2Q = \(\frac12\).(\(\frac11-\frac13\)) + \(\frac13\).(\(\frac12-\frac14\)) + ...+ \(\frac{1}{99}\).(\(\frac{1}{98}-\frac{1}{100}\))
2Q = \(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + ...+\(\frac{1}{98.99}\) - \(\frac{1}{99.100}\)
2Q = \(\frac12-\frac{1}{9900}\)
2Q = \(\frac{4949}{9900}\)
Q = \(\frac{4949}{9900}\) : 2
Q = \(\frac{4949}{19800}\)
Câu 1:
A = \(\frac14+\frac18+\frac{1}{16}+..+\frac{1}{128}\)
2A = \(\frac12+\frac14+\frac18+\cdots+\frac{1}{64}\)
2A - A = \(\frac12+\frac14+\frac18+\cdots+\frac{1}{64}\) - \(\frac14-\frac15-\frac{1}{16}-\ldots\frac{1}{128}\)
A = (\(\frac12-\frac{1}{128})+\left(\frac14-\frac14)+..+\left(\frac{1}{64}-\frac{1}{64}\right)\right.\)
A = \(\frac{64}{128}-\frac{1}{128}\) + 0 + 0+..+0
A = \(\frac{63}{128}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{99.100}=\frac{49}{99.100}\Rightarrow A=\frac{49}{2.99.100}\)