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Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(A=\frac{1}{2}.\left(\frac{4950-1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)
Ủng hộ mk nha!!
= 1/2.(2/1.2.3+2/2.3.4+.....+2/50.51.52
=1/2.(1/1.2-1/2.3+1/2.3-1/3.4+....+1/50.51-1/51.52
=1/2.(1/1.2-1/51.52)
=1/2.(1/2-1/2652)
=1/2.1325/2652
=1325/5304
A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/50.51-1/51.52
A=1/1.2-1/51.52
phần còn lại tự giải nhé
\(B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{49.50}\right)\)
Đến đây bạn tự tính nhé
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
\(B=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(B=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)
\(B=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{2450}\right)\)
\(B=\frac{1}{2}\cdot\frac{612}{1225}=\frac{306}{1225}\)
Vậy \(B=\frac{306}{1225}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{99.100}=\frac{49}{99.100}\Rightarrow A=\frac{49}{2.99.100}\)
\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)
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 = \(\frac{4949}{19800}\)