Rút gọn: S= (1- \(\frac{1}{1+2}\)) (1-\(\frac{1}{1+2+3}\))....(1- \(\frac{1}{1+2+3+.....+2016}\))
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31 tháng 1 2022
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\left(\dfrac{2015}{2}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}=\dfrac{1}{2017}\)
23 tháng 6 2017
\(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)
=\(\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}\)
=\(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
áp dụng vào biểu thức ta có\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
=\(1-\frac{1}{\sqrt{2016}}\)
đến đây cậu tự giải nốt nhé
8 tháng 5 2018
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=2018\)
8 tháng 5 2018
Ta có :
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=2018\)
Vậy \(A=2018\)
Chúc bạn học tốt ~
BN
6 tháng 5 2017
\(2C=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2016}}\)
\(2C-C=2+1+....+\frac{1}{2^{2016}}-\left(1+\frac{1}{2}+....+\frac{1}{2^{2017}}\right)\)
\(C\left(2-1\right)=2+1+....+\frac{1}{2^{2016}}-1-\frac{1}{2}-...-\frac{1}{2^{2017}}\)
\(C=2-\frac{1}{2^{2017}}=\frac{2^{2018}}{2^{2017}}-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
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Co cung ko cai dc
6 tháng 5 2017
\(C=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(2C=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(2C-C=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(C=2-\frac{1}{2^{2016}}\)
Với mọi \(n\in N\) ta có :
\(1-\frac{1}{1+2+3+...+n}=1-\frac{1}{\frac{n\left(n+1\right)}{2}}=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}\)
\(=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{n^2-n+2n-2}{n\left(n+1\right)}=\frac{n\left(n-1\right)+2\left(n-1\right)}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
Áp dụng ta được :
\(S=\frac{4.1}{2.3}.\frac{5.2}{3.4}......\frac{2018.2015}{2016.2017}\)
\(=\frac{\left(1.2.3....2015\right).\left(4.5....2018\right)}{\left(2.3.4.....2016\right).\left(3.4....2017\right)}=\frac{2018}{2016.3}=\frac{1009}{3024}\)