[toán 11] nhị thức Niuton

H

huy266

Ta có
[tex](2+1)^{2009}=C_{2009}^{0}2^{2009}+C_{2009}^{1}2^{2008}+...+C_{2009}^{2009}2^{0 }[/tex]
[tex](2-1)^{2009}=C_{2009}^{0}2^{2009}-C_{2009}^{1}2^{2008}+C_{2009}^{2}2^{2007}-C_{2009}^{2009}2^{0}[/tex]
[tex]S=\frac{3^{2009}-1}{2}[/tex]
 
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