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Question

$S=1.C^{0}_{2013} + 2C^{1}_{2013} + 3C^{2}_{2013} +...+2014C^{2013}_{2013}$

score 1 · Answer 1

Ta co

$C^0_{2113}=C^{2013}_{2013},C^1_{2013}=C^{2012}_{ 2013},C^2_{2013}=C^{2011}_{2013},...,C^{1006}_{2013}=C^{1007}_{2013}$

$\Rightarrow S=(1+2014)C^{0}_{2013}+(2+2013)C^1_{2013}+...+(1007+1008)C^{1006}_{2013}$

$S=2015(C^0_{2013}+C^1_{2013}+...+C^{1006}_{2013})$

$=2015.\frac{1}{2}(C^0_{2013}+C^1_{2013}+...+C^{1006}_{2013}+C^{1007}_{2013}+...+C^{2013}_{2013})$

$=\frac{2015}{2}.2^{2013}=2015.2^{2012}$

bạn ơi cách viết tổ hợp ntn vậy nhỉ?? – m2imlxx99 24-11-13 09:45 PM

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