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\(A=2014.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\right)\)
\(A=2014.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1007.2013}\right)\)
\(A=2.2014.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2014}\right)\)
\(A=2.2014.\frac{2013}{2014}\)
\(A=\frac{2.2014.2013}{2014}\)
\(A=2.2013\)
\(A=4026\)
xét mẫu(chỗ 1/2014 sửa lại thành 2/2014)
=(1/2015+1)+(2/2014+1)+...+(2013/3+1)+(2014/2+1)+(2015/1-2014)
=2016/2015+2016/2014+...+2016/3+2016/2+1
=2016.(1/2016+1/2015+...+1/4+1/3+1/2)
=> A= 1/2016
mún dễ hỉu hơn hãy gửi tin nhắn cho mik
gọi dãy số trên là A
ta có A<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
A<1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
A<1-\(\frac{1}{2014}\)=\(\frac{2013}{2014}\)
Vậy A < \(\frac{2013}{2014}\)
\(B=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2014}}\)
\(4B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)
\(3B=1-\frac{1}{4^{2014}}\)
\(B=\frac{1-\frac{1}{4^{2014}}}{3}\)
\(B=\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2013}}+\frac{1}{4^{2014}}\)
\(4B=1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2012}}+\frac{1}{4^{2013}}\)
\(4B-B=\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2012}}+\frac{1}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2013}}+\frac{1}{4^{2014}}\right)\)
\(3B=1-\frac{1}{4^{2014}}\)
\(B=\left(1-\frac{1}{4^{2014}}\right):3\)