Tính \(E=\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2012}}\)
K
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HT
10 tháng 10 2015
Xét mẫu:
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+2012}\)
= \(1+\frac{1}{3}+\frac{1}{6}+....+\frac{1}{2025078}\)
= \(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)
= \(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2012}-\frac{1}{2013}\right)\)
= \(1+2.\left(\frac{1}{2013}\right)\)
= \(\frac{4024}{2013}\)
=> E = \(\frac{2.2012}{\frac{4024}{2013}}\)
=> E = \(4024.\frac{2013}{4024}\)
=> E = 2013
29 tháng 8 2015
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2012}=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2.\frac{2012}{2013}\)\(\Rightarrow A=\frac{2.2012}{2.2012:2013}=\frac{1}{2013}\)
NC
3
26 tháng 9 2018
\(D=\frac{2.2012}{1+\frac{2}{2.\left(1+2\right)}+\frac{2}{2\left(1+2+3\right)}+\frac{2}{2\left(1+2+3+4\right)}+...+\frac{2}{2\left(1+2+..+2012\right)}}\)
\(=\frac{2.2012}{1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{4050156}}\)
\(=\frac{2.2012}{1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4050156}\right)}\)
\(=\frac{2.2012}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2012.2013}\right)}\)
\(=\frac{2.2012}{1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(=\frac{2.2012}{1+2.\left(\frac{1}{2}-\frac{1}{2013}\right)}\)
\(=\frac{2.2012}{1+\frac{2.2011}{2.2013}}\)
\(=\frac{2.2012}{1+\frac{2011}{2013}}\)
\(=\frac{4024}{\frac{4024}{2013}}\)
\(=2013\)
Vậy D=2013
NL
2
S
24 tháng 7 2018
Đặt A là tên biểu thức
Xét mẫu số, ta có: \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}\)
\(=1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}\)
\(=\frac{2}{2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}\)\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2\cdot\frac{2012}{2013}\)
\(\Rightarrow A=\frac{2.2012}{2\cdot\frac{2012}{2013}}=\frac{2012.2013}{2012}=2013\)
2 tháng 9 2021
\(\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}}\)
\(=\frac{2.2012}{1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}}\)
\(=\frac{2.2012}{\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}}\)
\(=\frac{2.2012}{2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(=\frac{2.2012}{2.\left(1-\frac{1}{2013}\right)}=\frac{2.2012}{2.\frac{2012}{2013}}=\frac{2012}{\frac{2012}{2013}}=\frac{2012.2013}{2012}=2013\)
JA
27 tháng 7 2016
Co quy luat nay ne em: 1+2=3=2.3:2; 1+2+3=6=3.4:2;...;1+2+3+...+2012=2012.2013:2
Suy ra ta co:
Mau so cua D=1 + 1/(2.3:2) + 1/(3.4:2) + 1/(4.5:2) + .... + 1/(2012.2013:2)
=1 + 2/2.3 + 2/3.4 + 2/4.5 + .... + 2/2012.2013
= 2.[1/2 + 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/2012.2013]
=2.[1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ..... + 1/2012.2013]
=2.[1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +....+1/2012 - 1/2013
=2[1 - 1/2013]
=2.2012/2013
Vay D= 2.2012 / (2.2012:2013)=2013
27 tháng 11 2016
Mẫu số = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}\)
\(=1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2012\right).2012:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(=2.\left(1-\frac{1}{2013}\right)=\frac{2.2012}{2013}\)
Phân số đề bài cho = \(\frac{2.2012}{\frac{2.2012}{2013}}=2013\)
MA
29 tháng 7 2015
phần mẫu số có
\(1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2012\right).2012:2}\)
\(1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{1}{2012.2013}\)
gọi tổng trên là S. lấy S : 2 có
\(S:2=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\)
\(S:2=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\)
\(S:2=\frac{1}{2}+\frac{1}{2}-\frac{1}{2013}\)
