1) Tính K + 2013 ,biết
K=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+......+\left(1+2+3+....+2013\right)}{2013x1+2012x2+2011x3+.....+2x2012+1x2013}\)
Những câu hỏi liên quan
Tính K + 2013, biết:
K=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2013\right)}{2013\times1+2012\times2+2011\times3+...+2\times2012+1\times2013}\)
k=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2013\right)}{2013\cdot1+2012\cdot2+2011\cdot3+...+2\cdot2012+1\cdot2013}\)
\(A=\frac{\left(1-2\right).\left(1+2\right)}{2^2}.\frac{\left(1-3\right).\left(1+3\right)}{3^2}.......\frac{\left(1-2013\right).\left(1+2013\right)}{2013^2}.\frac{\left(1-2014\right).\left(1+2014\right)}{2014^2}\)
Câu 2: K - 2013 = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2015\right)}{2015\times1+2014\times2+2013\times3+...+2\times2014+1\times2015}\)
Tìm K
\(K=\frac{1+\left(1+2\right)+...+\left(1+2+3+...+2013\right)}{2013.1+2012.2+2011.3+...+2.2012+1.2013}\)
Hỏi K + 2013 bằng bao nhiêu ?
Giúp mình với !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Tính A = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{2013}\left(1+2+...+2013\right)\)
theo công thức \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)
=>\(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{2013}.\frac{2013.2014}{2}\)
\(=>A=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2014}{2}=>A=\frac{1}{2}\left(1+2+3+..+2014\right)-\frac{1}{2}\)
\(=>A=\frac{1}{2}.\frac{2014.2015}{2}-\frac{1}{2}=1014552\)
k=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2013\right)}{2013\cdot1+2012\cdot2+2011\cdot3+...+2\cdot2012+1\cdot2013}\)
K+2003=?
(Các bạn hãy giúp mình trả lời giúp mình bài này mau nhé, bạn nào đúng mình sẽ tick cho)
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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\(K=\frac{1+\left(1+2\right)+...+\left(1+2+...+2013\right)}{2013.1+2012.2+...+2.2012+1.2013}\)
Hỏi K + 2013 = ?
Giúp mk nha !
Mk nghĩ ra rồi :
\(K=\frac{1+\left(1+2\right)+...+\left(1+2+...+2013\right)}{2013.1+2012.2+2011.3+...+1.2013}\)
Ta thấy có 2013 số 1 ở tử số, 2012 chữ số 2, ..., vậy ta có :
\(K=\frac{1.2013+2.2012+...+2013.1}{2013.1+2012.2+...+1.2013}\)
\(\Rightarrow K=1\)\(\Rightarrow K+2013=2014\)
Đ/S : 2014
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