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A=1/2.9+1/9.7+1/7.19+...+1/252.509
=?
??????
b: Ta có: \(B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{65}{4\cdot69}\)
\(=\dfrac{13}{276}\)
\(A=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+...+\dfrac{2}{97\cdot100}\\ A=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{2}{3}\cdot\dfrac{99}{100}=\dfrac{33}{50}\\ B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\\ B=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{69}\right)=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)
\(C=70\left(\dfrac{13}{56}+\dfrac{13}{72}+\dfrac{13}{90}\right)=70\cdot13\left(\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ C=910\left(\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{10}\right)=910\cdot\dfrac{3}{70}=39\)
b)\(\dfrac{1}{7}B=\dfrac{1}{10.18}+\dfrac{1}{18.26}+\dfrac{1}{26.34}+...+\dfrac{1}{802.810}\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{8}{10.18}+\dfrac{8}{18.26}+\dfrac{8}{26.34}+...+\dfrac{8}{802.810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{34}+...+\dfrac{1}{802}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}.\dfrac{8}{81}\)
\(\dfrac{1}{7}B=\dfrac{1.8}{8.81}\)
\(\dfrac{1}{7}B=\dfrac{1}{81}\)
\(B=\dfrac{1}{81}:\dfrac{1}{7}\)
\(B=\dfrac{7}{81}\)
a: \(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{121}-\dfrac{1}{124}=1-\dfrac{1}{124}=\dfrac{123}{124}\)
b: \(=3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)=3\cdot\dfrac{99}{202}=\dfrac{297}{202}\)
c: \(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{401}-\dfrac{1}{405}\right)=\dfrac{1}{4}\cdot\dfrac{404}{405}=\dfrac{101}{405}\)
d: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
Đặt \(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{9}{196}.\dfrac{1-3-5-7-...-49}{89}\)
Đặt \(B=1-3-5-7-..-49\)
\(=1-\left(3+5+7+...+49\right)\)
\(=1-\left\{\left(49+3\right).\left[\left(49-3\right):2+1\right]:2\right\}\)
\(=1-624\)
\(=-623\)
\(\Rightarrow\dfrac{9}{196}.\left(\dfrac{-623}{89}\right)=-\dfrac{9}{28}\)
Vậy: \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}=-\dfrac{9}{28}\)
Xét \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)
=\(\dfrac{1}{5}.\dfrac{45}{196}\)
=\(\dfrac{9}{196}\)
Xét \(\dfrac{1-3-5-7-..-49}{89}\)
=\(\dfrac{1-\left(3+5+7+...+49\right)}{89}\)
CT tính sl số hạng (số cuối - số đầu ):2+1
số lượng số hạn của dãy 3+5+7+...+49 là (49-3):2+1=24
Áp dụng CT tính tổng số hạng dãy số cách đều Tổng = [ (số đầu + số cuối) x Số lượng số hạng ] : 2
=> tổng = [(3+49).24]:2=624
=>\(\dfrac{1-624}{89}\)
=\(\dfrac{-623}{89}\)
=-7
từ đó ta có \(\dfrac{9}{196}.\left(-7\right)=\dfrac{-9}{28}\)
Ta có: \(A=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{5}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{49-4}{4\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
\(\Leftrightarrow A=\dfrac{1}{5}\cdot\dfrac{45}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)
\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)
\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{-623}{89}=-\dfrac{9}{28}\)
b: \(C=75\left(2-128+128\right)=75\cdot2=150\)
e: \(E=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{69\cdot74}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{69}-\dfrac{1}{74}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{70}{74}=\dfrac{14}{74}=\dfrac{7}{37}\)
