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A=4/3+9/8+16/15+..............+4064256/4064255
A=1+1/3+1+1/8+1/15+...............+1/4064255
A=(1+1+...+1)+(1/3+1/8+...+1/406255) (có 2015 số 1)
A=2015+(1/1.3+1/2.4+...........+1/2015.2017)
A=2015+1/2(1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+....+1/2012-1/2014+1/2013-1/2015+1/2014-1/2016+1/2015-1/2017)
A=2015+1/2(1+1/2-1/2016-1/2017)
A=2015,749504
k cho mình nhé mình k lại cho
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.,,\frac{50^2}{49.51}\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.,,\frac{50.50}{49.51}\)
=\(\frac{\left(2.3.4...50\right).\left(2.3.4...50\right)}{\left(1.2.3....49\right).\left(3.4.5....51\right)}\)
=\(\frac{50.2}{1.51}\)
=\(\frac{100}{51}\)
\(=\frac{2.3.4...50}{1.2.3...49}.\frac{2.3.4...50}{3.4.5...51}=50.\frac{2}{51}=\frac{100}{51}\)
a) A = \(\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+....+\frac{10301}{100.103}\) (có 34 số hạng)
A = \(\frac{4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+....+\frac{100.103+1}{103.100}\)
A = \(1+\frac{1}{1.4}+1+\frac{1}{4.7}+1+\frac{1}{7.10}+....+1+\frac{1}{100.103}\)
A = \(1.34+\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}\cdot\frac{102}{103}\)
A = \(34+\frac{34}{103}=\frac{3536}{103}\)
\(\frac{-2}{1.3}-\frac{2}{3.5}-...-\frac{2}{13.15}\)
\(=-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}\right)\)
\(=-\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=-\left(1-\frac{1}{15}\right)\)
\(=\frac{-14}{15}\)
\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)
\(B=\frac{\left(2.3.4...150\right)\left(2.3.4...150\right)}{\left(1.2.3...149\right)\left(3.4.5...151\right)}\)
\(B=\frac{\left(1.2.3...149\right).150.2.\left(3.4.5...150\right)}{\left(1.2.3...149\right).\left(3.4.5...150\right).151}\)
\(B=\frac{300}{151}\)
có thể giải kĩ hơn ko ?
Ơ mình làm chi tiết rồi mà nhỉ ??
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{150^2}{149.151}\)
\(=\frac{2.2.3.3.4.4.....150.150}{1.3.2.4.3.5.....148.150.149.151}\)
\(=\frac{2.3.4....150.2.3.4....150}{2.3.4....148.149.3.4.5.6.....150.151}=\frac{150.2}{151}=\frac{300}{151}\)
Bài làm:
Ta có: \(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{150^2}{149.151}\)
\(B=\frac{2^2.3^2.4^2.....150^2}{\left(1.2.3.....149\right)\left(3.4.5.....151\right)}\)
\(B=\frac{2^2.3^2.4^2.....150^2}{1.2.3^2.4^2.....149^2.150.151}\)
\(B=\frac{2.150}{151}\)
\(B=\frac{300}{151}\)
\(B=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}\cdot\frac{4^2}{3.5}\cdot....\cdot\frac{150^2}{149.151}\)
\(B=\frac{2^2\cdot3^2\cdot4^2\cdot...\cdot150^2}{\left(1\cdot2\cdot3\cdot....\cdot149\right).\left(3\cdot4\cdot5\cdot...\cdot151\right)}\)
\(B=\frac{2^2.3^2.4^2....150^2}{1.2.3^2.4^2....149^2.150.151}\)
\(B=\frac{2.150}{151}\)
\(B=\frac{300}{151}\)
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