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\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{2017}.\left(1+2+3+...+2017\right)\)
\(S=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{2017}.\frac{\left(1+2017\right).2017}{2}\)
\(S=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2018}{2}\)
\(S=\frac{1}{2}.\left(2+3+4+...+2018\right)\)
\(S=\frac{1}{2}.\frac{\left(2+2018\right).2017}{2}\)
\(S=\frac{2020.2017}{4}=505.2017=1018585\)
\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2017}\)
\(S=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2035153}\)
\(S=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{4070306}\)
\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2017.2018}\)
\(S=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{2018}\right)=2.\frac{504}{1009}=\frac{1008}{1009}\)
Vậy \(S=\frac{1008}{1009}\)
P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
Ta có: \(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)
\(=1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)\)
\(=\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2018}\)
\(=2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)
Giờ ta thế vào bài toán ban đầu được
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
\(=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}\)
\(=\frac{2017}{2018}\)
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
1. Bài giải:
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\)
\(\Rightarrow\frac{1}{2}A=A-\frac{1}{2}A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1000}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\right)\)
\(\Rightarrow\frac{1}{2}A=1-\frac{1}{1002}=\frac{1001}{1002}\Rightarrow A=\frac{2002}{1002}=\frac{1001}{501}\)
Vậy \(A=\frac{1001}{501}\)