1\(^2\)+2\(^2\)+(\(\frac{1}{2}\))=???
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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 \(A=\frac{4052168}{2013}\)
S
27 tháng 7 2019
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+.....-\frac{1}{2^{99}}\Rightarrow2A+A=3A=\left(1-\frac{1}{2}+\frac{1}{2^2}-....-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+......-\frac{1}{2^{100}}\right)=1-\frac{1}{2^{100}}=\frac{2^{100}-1}{2^{100}}\Rightarrow A=\frac{2^{100}-1}{3.2^{100}}\)
\(2,4B=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\Rightarrow4B-B=3B=\left(2+\frac{1}{2}+....+\frac{1}{2^{97}}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)=2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\Rightarrow B=\frac{2^{100}-1}{3.2^{99}}\)
\(3,C=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\Rightarrow8C=4-\frac{1}{2}+\frac{1}{2^4}-.....-\frac{1}{2^{55}}\Rightarrow8C+C=9C=\left(4-\frac{1}{2}+\frac{1}{2^4}-....-\frac{1}{2^{55}}\right)+\left(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\right)=4-\frac{1}{2^{58}}=\frac{2^{60}-1}{2^{58}}\Rightarrow C=\frac{2^{60}-1}{9.2^{58}}\)
14 tháng 7 2017
\(F=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(F=\left(\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-2.\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{50}}\right)\)
\(F=\frac{1}{2^{51}}+\frac{1}{2^{52}}+...+\frac{1}{2^{100}}\)
14 tháng 7 2017
\(E=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2E=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2E-E=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(E=1-\frac{1}{2^{100}}\)
NM
20 tháng 8 2017
Ta có:
\(\sqrt{\frac{1}{1^2}+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^4+2n^3+3n^2+2n+1}{n^2.\left(n+1\right)^2}}\)
\(=\sqrt{\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(N+1\right)}=1+\frac{1}{n\left(n+1\right)}\)
\(=1+\frac{1}{n}-\frac{1}{n+1}\)
Thế vào bài toán ta được
\(S=1+1+...+1+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=98+\frac{1}{2}-\frac{1}{100}=\frac{9849}{100}\)
1^2+2^2+1/2= 1+4+0,5=5,5
12 + 22 + \(\frac{1}{2}\)
= 1 + 4 + \(\frac{1}{2}\)
= 5 + \(\frac{1}{2}\)
=\(\frac{2.5}{2}\)+ \(\frac{1}{2}\)
= \(\frac{11}{2}\)