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Mình làm được rồi này :
\(B=\frac{1}{1.2.3}-\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{97.98.99}\right)\)
\(=\frac{1}{6}-\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{97.98}-\frac{1}{98.99}\right)\)
\(=\frac{1}{6}-\left(\frac{1}{2.3}-\frac{1}{98.99}\right)\)
\(=\frac{1}{6}-\frac{1}{6}+\frac{1}{9702}\)
\(=\frac{1}{9702}\)
Ta có:
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{998.999.1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{998.999.1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{998.999}-\frac{1}{999.1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{999.1000}=\frac{499499}{999000}\Leftrightarrow A=\frac{499499}{1998000}\)
\(B=\frac{1}{1.2.3.4.5}+\frac{1}{2.3.4.5.6}+\frac{1}{3.4.5.6.7}+\frac{1}{996.997.998.999.1000}\)
\(\Rightarrow\frac{1}{4}B=\frac{4}{1.2.3.4.5}+\frac{4}{2.3.4.5.6}+\frac{4}{3.4.5.6.7}+....+\frac{4}{996.997.998.999.1000}\)
\(\Rightarrow\frac{1}{4}B=\frac{1}{1.2.3.4}-\frac{1}{2.3.4.5}+\frac{1}{2.3.4.5}-\frac{1}{3.4.5.6}+\frac{1}{3.4.5.6}-\frac{1}{4.5.6.7}+...+\frac{1}{996.997.998.999}-\frac{1}{997.998.999.1000}\)
\(\Rightarrow\frac{1}{4}B=\frac{1}{1.2.3.4}-\frac{1}{997.998.999.1000}=\frac{41417124749}{994010994000}\Leftrightarrow B=\frac{41417124749}{3976043976000}\)
A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
= \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
= \(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)\)
= \(\frac{1}{2}.\frac{390}{781}=\frac{195}{781}\)
bạn có: 3G = (5 + 8/3) + (11/3^2 + 14/3^3 + ... + 299/3^98 + 302/3^99)
G = 5/3 + (8/3^2 + 11/3^3 + .... + 296/3^98 + 299/3^99 + 302/3^100)
bạn có 3G - G = 5 + 8/3 - 5/3 + (11/3^2 - 8/3^2) + (14/3^3 - 11/3^3) + .... + (299/3^98 - 296/3^98) + (302/3^99 - 299/3^99) - 302/3^100
hay 2G = 5 +8/3 - 5/3 + (3/3^2 + 3/3^3 + ... + 3/3^98 + 3/3^99) - 302/3^100
2G = 6 + (1/3 + 1/3^2 +... + 1/3^97 + 1/3^98)
đặt H = 1/3 + 1/3^2 + ... + 1/3^97 + 1/3^98
suy ra ta có 3H = 1 + 1/3 + .... + 1/3^96 + 1/3^97
3H - H = 1 - 1/3^98 hay 2H = 1 - 1/3^98
ở trên bạn có:
2G = 6 + (1/3 + 1/3^2 +... + 1/3^97 + 1/3^98)
hay 2G = 6 + H
hay 4G = 12 + 2H
hay 4G = 12 + 1 - 1/3^98
hay G = 13/4 - (1/3^98)/4
tìm được giá trị của G rồi thì bạn dễ dàng tìm được các bước tiếp theo thôi :D, sr vì tớ lười :D
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(-5\right)\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(=-\frac{3}{5}\)
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(\frac{-5}{8}\right)-\left(\frac{-5}{10}\right)+\left(\frac{-5}{11}\right)+\left(\frac{-5}{12}\right)}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(-5\right).\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(=\frac{-3}{5}\)