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Đặt \(A=2^2+2^2+2^3+...+2^{2013}\)
\(\Leftrightarrow2A=2^3+2^3+2^4+...+2^{2014}\)
\(\Leftrightarrow A=2^{2014}\)
Theo đề, ta có: \(2^x=2^{2014}\)
hay x=2014
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câu 1: -799999
câu 2: cần 13245 chữ số
câu 3: 2014 chữ số
câu 4: -617
câu 6: 2014
câu 7: 16
câu 10: 9
Còn mấy câu nữa mình không biết. bạn tích đúng cho mình nha
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=> B=2013. (1+\(\frac{1}{1+2}\) +\(\frac{1}{1+2+3}\) +...+ \(\frac{1}{1+2+3+...+2012}\))
=>B= 2013.(\(\frac{2}{2}\) + \(\frac{2}{2.3}\) +\(\frac{2}{3.4}\) +...+\(\frac{2}{2012.2013}\))
=>B= 2013.2.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +\(\frac{1}{3.4}\) +...+\(\frac{1}{2012.2013}\))
=>B=4026. (1-\(\frac{1}{2}\) +\(\frac{1}{2}\) -\(\frac{1}{3}\) + ...+\(\frac{1}{2012}\) - \(\frac{1}{2013}\))
=>B=4026.(1-\(\frac{1}{2013}\))
=>B=4026.\(\frac{2012}{2013}\) => B=2.2012=4024 Vậy B=4024
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Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
Lời giải:
Đặt $A=2^2+2^2+2^3+2^4+...+2^{2013}$
$2A=2^3+2^3+2^4+2^5+...+2^{2014}$
$\Rightarrow A=2A-A=(2^3+2^3+2^4+2^5+...+2^{2014})-(2^2+2^2+2^3+2^4+...+2^{2013})$
$\Rightarrow A=(2^3+2^{2014})-(2^2+2^2)=2^{2014}$
Vậy $A=2^{2014}=2^x$
$\Rightarrow x=2014$
x=2014