tính tổng:
\(s=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\)
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31 tháng 12 2017 lúc 22:12
Tính tổng :
\(S=\frac{1}{2013-1}+\frac{2}{2013+1}+\frac{2^2}{2013^2+1}+\frac{2^3}{2013^{2^2}+1}+.....+\frac{2^{n+1}}{2013^{2^n}+1}\)
tính tổng S=\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{2012^2}+\frac{1}{2013^2}}\)
giúp mình với mơn nha
br258 / 6.18 dư 3 , khi chia 12 ,3 , 21 dư 6 vậy br = 26 .1 / 655
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Tính tổng \(M=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+...+\frac{2013}{1+2013^2+2014^2}\)
bn ơi mik nhớ, bn ơi mik rất nhớ cái tick
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tính tổng \(S=\frac{1}{2^{-2013}+1}+\frac{1}{2^{-2012}+1}+...+\frac{1}{2^0+1}+...+\frac{1}{2^{2012}+1}+\frac{1}{2^{2013}+1}\)
\(S=\frac{2^{2013}}{2^{2013}+1}+\frac{2^{2012}}{2^{2012}+1}+....+\frac{1}{2^{2012}+1}+\frac{1}{2^{2013}+1}\)
=(\(\frac{2^{2013}}{2^{2013}+1}+\frac{1}{2^{2013}+1}\))+(\(\frac{2^{2012}}{2^{2012}+1}+\frac{1}{2^{2012}+1}\))+...+ \(\frac{1}{2}\) ( có 2013 dấu ngoặc )
= 1+ 1+.....+ \(\frac{1}{2}\) = 2013\(\frac{1}{2}\)
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Tính tổng: \(A=\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+\frac{1}{4.\left(1+2+3+4\right)}+...+\frac{1}{2013.\left(1+2+3+...+2013\right)}\)
Lời giải:
** Sửa đề:
$A=\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+\frac{1}{4}(1+2+3+4)+....+\frac{1}{2013}(1+2+3+...+2013)$
$A=\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+....+\frac{1}{2013}.\frac{2013.2014}{2}$
$=\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{2014}{2}$
$=\frac{3+4+5+...+2014}{2}$
$=\frac{1+2+3+4+5+...+2014}{2}-\frac{3}{2}$
$=\frac{2014.2015:2}{2}-\frac{3}{2}$
$=1014551$
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Câu 1: Rút gọn: \(A=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)\)
Câu 2: Cho \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\). Tính \(\left(S-P\right)^{2013}\)
Tính \(A=2013+\frac{2013}{1+2}+\frac{2013}{1+2+3}+\frac{2013}{1+2+3+4}+...+\frac{2013}{1+2+3+...+2012}\)
Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
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Tính giá trị biểu thức \(S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}}{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{2}{2013}+\frac{1}{2014}}\) .
\(S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}}\)
Xét mẫu:
\(\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}\)
= \(\left(1+\frac{2013}{2}\right)+\left(1+\frac{2012}{3}\right)+...+\left(1+\frac{1}{2014}\right)+1\)
= \(\frac{2014}{2}+\frac{2014}{3}+....+\frac{2014}{2013}+\frac{2014}{2014}\)
= \(2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(\Rightarrow S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}\)
\(\Rightarrow S=\frac{1}{2014}\)
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Tính giá trị biểu thức \(S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}}{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{2}{2013}+\frac{1}{2014}}\) .