Tính: \(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2017}\left(1+2+3+...+2017\right)\)
K
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Những câu hỏi liên quan
DL
25 tháng 6 2019
Ta có: \(A=\frac{\left(1+\frac{2017}{1}\right)\left(1+\frac{2017}{2}\right)...\left(1+\frac{2017}{1009}\right)}{\left(1+\frac{1009}{1}\right)\left(1+\frac{1009}{2}\right)...\left(1+\frac{1009}{2017}\right)}=\frac{\frac{2017+1}{1}\frac{2017+2}{2}...\frac{2017+1009}{1009}}{\frac{1009+1}{1}\frac{1009+2}{2}...\frac{1009+2017}{2017}}\)
\(\Leftrightarrow A=\frac{\frac{2018.2019...3026}{1.2...1009}}{\frac{1010.1011...3026}{1.2...2017}}=\frac{2018.2019...3026}{1.2...1009}.\frac{1.2...2017}{1010.1011...3026}\)
\(\Leftrightarrow A=\frac{1.2...2017.2018.2019...3026}{1.2...1009.1010.1011...3026}=\frac{1.2.3...3026}{1.2.3...3026}=1.\)
S
12 tháng 7 2018
a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)
b, c cùng 1 câu phải k
ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)
LD
15 tháng 7 2018
A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)
NHA
HỌC TỐT
AN
5 tháng 3 2019
\(\Rightarrow C=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)....\left(1-\frac{1}{1+2++...+2017}\right)\)
\(\Rightarrow C=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)....\left(1-\frac{1+2+...+2017-1}{1+2+3+...+2017}\right)\)
\(\Rightarrow C=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)....\left(\frac{2017.2018:2-1}{2017.2018:2}\right)\)
\(\Rightarrow C=\frac{2}{3}.\frac{5}{6}...\frac{2017.2018:2-1}{2017.2018:2}\)
\(\Rightarrow C=\frac{4}{6}.\frac{10}{12}....\frac{2017.2018-1}{2017.2018}\)
\(\Rightarrow C=\frac{1.4}{2.3}.\frac{2.5}{3.4}....\frac{2016.2019}{2017.2018}\)
\(\Rightarrow C=\frac{1.2....2016}{2.3.....2018}.\frac{4.5....2019}{3.4....2017}\)
\(\Rightarrow C=\frac{1}{2017.2018}.\frac{2018.2019}{3}\)
\(\Rightarrow C=\frac{673}{2017}\)
5 tháng 3 2019
đầu bài mình đặt là A mà giải lại là C =.= nhưng mà ko sao vì bạn làm đúng rồi. cảm ơn An Nguyễn nhé <3
4 tháng 9 2017
\(A=\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+\frac{6}{2}+..........+\frac{2019}{2}=\frac{3+4+5+..............+2019}{2}.\)
ta có 3+4+5+......+2019=(3+2019)2016:2=2038176
=>\(\frac{2038176}{2}=1019088\)
NN
26 tháng 3 2017
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right)\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{2016.2018}{2017^2}\)
\(=\frac{2.3^2.4^2.5^2...2016^2.2017.2018}{2^2.3^2.4^2.5^2...2017^2}\)
\(=\frac{2018}{2.2017}=\frac{1009}{2017}\)
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2017}\left(1+2+...+2017\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{2017}.\frac{2017\left(2017+1\right)}{2}\)
\(=1+\frac{2.3}{2.2}+\frac{3.4}{3.2}+....+\frac{2017.2018}{2017.2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2018}{2}\)
\(=\frac{2+3+4+...+2018}{2}\)
\(=\frac{\frac{2018\left(2018+1\right)}{2}-1}{2}\)
\(=1018585\)
Suy ra A=1+1.5+2+....+1009=1 013 532.5