[Toán 11] Niu Tơn

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nguyenbahiep1

Chứng minh:

$1.C_{2010}^0 + 3^2. C_{2010}^2+...+3^{2010}.C_{2010}^2010= 2^{2009}. (2^{2010} + 1)$


[laTEX] (1-3)^{2010} = (-2)^{2010} = C_{2010}^0 + C_{2010}^2.3^2 + C_{2010}^4.3^4 + ........+ C_{2010}^{2010}.3^{2010} - ( 3^1.C_{2010}^1 + C_{2010}^3.3^3 + C_{2010}^5.3^5 + ........+ C_{2010}^{2009}.3^{2009}) = A -B \\ \\ (1+3)^{2010} = (4)^{2010} = C_{2010}^0 + C_{2010}^2.3^2 + C_{2010}^4.3^4 + ........+ C_{2010}^{2010}.3^{2010} + ( 3^1.C_{2010}^1 + C_{2010}^3.3^3 + C_{2010}^5.3^5 + ........+ C_{2010}^{2009}.3^{2009}) = A +B \\ \\ \begin{cases} A+B = 4^{2010} \\ A-B = 2^{2010} \end{cases} \\ \\ 2.A = 4^{2010}+2^{2010} \\ \\ A = 2^{4019} + 2^{2009} = 2^{2009}.(2^{2010} +1) \Rightarrow dpcm[/laTEX]
 
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