Tính giá trị biểu thức
A=\(\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+...+\frac{2013}{1+2013^2+2013^4}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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(\(1\frac{1}{2013}=2013.2.\frac{2012}{2013}=4024\)
13 tháng 8 2017
NX \(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}\) =\(\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}-1\right)}{\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}+1\right)^2}\)
=\(\frac{\left(\left(\sqrt{n+1}-\sqrt{n}\right)^2-1^2\right)}{n+1-n-1-2\sqrt{n}}\) \(=\frac{n+1+n-2\sqrt{\left(n+1\right)n}-1}{-2\sqrt{n}}=\frac{2n-2\sqrt{n\left(n+1\right)}}{-2\sqrt{n}}\)
=\(\frac{n-\sqrt{n\left(n+1\right)}}{-\sqrt{n}}=\frac{n}{-\sqrt{n}}+\frac{\sqrt{n\left(n+1\right)}}{\sqrt{n}}=-\sqrt{n}+\sqrt{n+1}\)
thay vao Q ta co
Q= \(-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{2012}+\sqrt{2013}=-\sqrt{2}+\sqrt{2013}\)
CH
Cô Hoàng Huyền
Admin
VIP
8 tháng 5 2017
Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
4 tháng 5 2016
=> 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
Xét \(\frac{n}{1+n^2+n^4}=\frac{n}{n^4+2n^2+1-n^2}=\frac{n}{\left(n^2+1\right)^2-n^2}=\frac{n}{\left(n^2-n+1\right)\left(n^2+n+1\right)}=\frac{1}{2}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{2013^2-2013+1}-\frac{1}{2013^2+2013+1}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{2013^2+2013+1}\right)=...\)