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\(A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\)
\(2A=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
\(2A-A=\left(2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)-\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\right)\)
\(A=1+\frac{1}{2}-\frac{1}{2^{2014}}\)
\(A=\frac{2^{2014}}{2^{2014}}+\frac{2^{2013}}{2^{2014}}-\frac{1}{2^{2014}}\)
\(A=\frac{2^{2014}+2^{2013}-1}{2^{2014}}\)
Tham khảo nhé~
sao lại lấy ảnh của tui.
bài cậu hỏi tôi làm rồi đó
nhớ ****
Áp dụng công thức:
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\) thì được
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2009}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2009.2010}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2009.2010}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{1004}{1005}\)
thôi, làm luôn nè
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2009}\)
\(=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+2009\right).2009:2}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2009.2010}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+2.\left(\frac{1}{4}-\frac{1}{5}\right)+...+2.\left(\frac{1}{2009}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
a) -1 - 2 + 3 + 4 - 5 - 6 + 7 + 8 - 9 - 10 + 11 + 12 - ... - 2013 - 2014 + 2015 + 2016
= ( -1 - 2 + 3 + 4 ) - ( 5 + 6 - 7 - 8 ) - ( 9 + 10 - 11 - 12 ) - .......... - ( 2013 + 2014 - 2015 - 2016 )
= 4 - ( -4 ) - ( -4 ) - ......... - ( -4 )
= 4 + 4 + 4 +....... + 4
= { [ ( 2016 - 1 ) : 1 + 1 ] : 4 } . 4
= { [ 2015 : 1 + 1 ] : 4 } . 4
= { 2016 : 4 } . 4
= 504 . 4
= 2016
b) \(\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):\left(\frac{1}{4}-1\right):\left(\frac{1}{5}-1\right):.........:\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}:\frac{-2}{3}:\frac{-3}{4}:\frac{-4}{5}:......:\frac{-99}{100}\)
\(=\frac{-1}{2}.\frac{3}{-2}.\frac{4}{-3}.\frac{5}{-4}.......\frac{100}{-99}\)
\(=\frac{-1.3.4........100}{2.2.3.4......99}\)
\(=\frac{-1.100}{2.2}\)
\(=\frac{-100}{4}\)
\(=-25\)
a) -1-2+3+4-5-6+7+8+...+2016=-3+3-7+7-...-2016+2016=0
b) \(\left(\frac{1}{2}-1\right):...:\left(\frac{1}{100}-1\right)=\frac{-1}{2}:\frac{-2}{3}:\frac{-3}{4}:...:\frac{-99}{100}\)
\(=\)\(\frac{-1}{2}.\frac{-3}{2}.....\frac{-100}{99}=\frac{-1}{2}.\left(-50\right)=25\)
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(\(1\frac{1}{2013}=2013.2.\frac{2012}{2013}=4024\)
Xét mẫu số của A :
\(\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\) \(=\left(1+1+1+...+1\right)+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
\(=\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1\)\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}\)\(=2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
Mẫu số gấp 2014 lần tử số nên A = \(\frac{1}{2014}\)
A =\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+...+\(\frac{1}{1+2+3+4...+2014}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\)
\(\Rightarrow2A=2\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\right)\)
\(\Rightarrow2A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4058210}\)
\(\Rightarrow2A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{2015}\)
\(\Rightarrow2A=\frac{2013}{4030}\)
\(\Rightarrow A=\frac{2013}{8060}\)
ngài Kiệt ღ ๖ۣۜLý๖ۣۜ đúng là không ái sánh bằng sự gian xảo này