B\(\dfrac{5}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)ài 1:
S1 : \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{97.100}\)
= \(\dfrac{5}{3}.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}\right)\)
= \(\dfrac{5}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
= \(\dfrac{5}{3}.\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{5}{3}.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
= \(\dfrac{5}{3}.\dfrac{99}{100}\)
= \(\dfrac{33}{20}\)
S3 : \(\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)
= \(\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)
= \(\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)
= \(\dfrac{5}{3}.\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{25.28}\right)\)
= \(\dfrac{5}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
=\(\dfrac{5}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
= \(\dfrac{5}{3}.\left(\dfrac{7}{28}-\dfrac{1}{28}\right)\)
= \(\dfrac{5}{3}.\dfrac{6}{28}\)
= \(\dfrac{5}{3}.\dfrac{3}{14}\)
= \(\dfrac{5}{14}\)
S2 : \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
= \(\dfrac{1}{2}.\dfrac{16}{51}\)
= \(\dfrac{8}{51}\)
S4 : \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{300}}\)
3S = \(\dfrac{3}{3}+\dfrac{3}{3^2}+\dfrac{3}{3^3}+\dfrac{3}{3^4}+...+\dfrac{3}{3^{300}}\)
3S = \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{299}}\)
3S-S = \(\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{299}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{300}}\right)\)
2S = \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{299}}-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-\dfrac{1}{3^4}-...-\dfrac{1}{3^{300}}\)
2S = \(1-\dfrac{1}{3^{300}}\)
