tính: \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
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24 tháng 9 2016
Mẫu số của A \(=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\)
\(=\left(1+1+...+1\right)+\left(\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\right)\)
(2012 số 1) (2011 phân số)
\(=\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1\)
\(=\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}\)
\(=2013.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)
=> \(A=\frac{1}{2013}\)
24 tháng 9 2016
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)}\)
\(\Rightarrow A=\frac{1}{2013}\)
Vậy \(A=\frac{1}{2013}\)
IL
6
19 tháng 5 2019
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\)
\(A-\frac{1}{2}A=1-\frac{1}{2^{2013}}\)
\(A=\frac{1-\frac{1}{2^{2013}}}{\frac{1}{2}}\)
19 tháng 5 2019
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}+\frac{1}{2^{2013}}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(=\frac{1}{2^{2013}}-1\)
mik ko chắc lắm(đừng ném đá mik nha)
#sadgirl#
LN
2
18 tháng 4 2017
A=1+1/2+1/2^2+...+1/2^2012
2A=2(1+1/2+1/2^2+...+1/2^2012
2A=2+1+1/2+1/2^2+...+1/2^2011
2A-A=2+[(1+1/2+1/2^2+...+1/2^2011)-(1+1/2+1/2^2+...+1/2^2011)]-1/2^2012 (mình làm tắt 1 bước)
A=2-1/2^2012 ! CÒN LẠI TỰ TÍNH
D
18 tháng 4 2017
A=\(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
2A=\(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
2A-A=A=\(\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
A=\(2-\frac{1}{2^{2012}}\)
11 tháng 7 2016
\(\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2013}{1}+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}-2012}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2013}{1}-1\right)+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4024}{2012}-1\right)}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2012}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)}\)
\(=\frac{1}{2012}\)
Ủng hộ mk nha ^_-
3 tháng 3 2016
TA TÁCH 2012 RA THÀNH 2012 CON SỐ 1.LẤY (1 + 2012/2) + (1 + 2011/3) + (1 + 2010/4); +...+ (1 + 1/2013) Ở MẪU, TA ĐƯỢC 2014/2 + 2014/3 +...+ 2014/2013(Ở MẪU).ĐẶT THỪA SỐ CHUNG 2014 RA NGOÀI TA SẼ ĐƯỢC 2014(1/2 + 1/3 +...+ 1/2013)(Ở MẪU).LẤY TỬ CHIA MẪU TA SẼ CÒN LẠI 1/2014. VẬY A=1/2014
8 tháng 9 2015
Xét \(P=\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\) với a>0
\(P^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}\)
\(=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\left(\frac{a^2+a+1}{a\left(a+1\right)}\right)^2\)
Do a>o nên \(P=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng kết quả của P ta có:
\(A=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}+\frac{1}{3}\right)+....+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\) \(A=2012+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=2012+1-\frac{1}{2013}\)
\(A=2013-\frac{1}{2013}=\frac{4052168}{2013}\)
Vậy \(A=\frac{4052168}{2013}\)
8 tháng 1 2017
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
29 tháng 8 2015
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2012}=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2.\frac{2012}{2013}\)\(\Rightarrow A=\frac{2.2012}{2.2012:2013}=\frac{1}{2013}\)
\(\Rightarrow2A=2\left(1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(-\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
\(A=2-\frac{1}{2^{2012}}\)
Vậy \(A=2-\frac{1}{2^{2012}}\)