tính giá trị biểu thức
\(A=\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}.....+\frac{1}{1+2+......2012}}\)
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29 tháng 8 2015
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2012}=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2.\frac{2012}{2013}\)\(\Rightarrow A=\frac{2.2012}{2.2012:2013}=\frac{1}{2013}\)
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
\(\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}}\)
\(=\frac{2.2012}{1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}}\)
\(=\frac{2.2012}{\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}}\)
\(=\frac{2.2012}{2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(=\frac{2.2012}{2.\left(1-\frac{1}{2013}\right)}=\frac{2.2012}{2.\frac{2012}{2013}}=\frac{2012}{\frac{2012}{2013}}=\frac{2012.2013}{2012}=2013\)
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
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 = \(\frac{2.2012}{\frac{2.2012}{2013}}=2013\)
19 tháng 3 2018
B=2013.(1+
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{2013}{1+2+3+...+2012}\)
B=2013(\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)
B=2013.2(\(1\frac{1}{2013}=2013.2.\frac{2012}{2013}=4024\)
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
\(S:2=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\)
\(S:2=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\)
\(S:2=\frac{1}{2}+\frac{1}{2}-\frac{1}{2013}\)