nhị thức newton khó

N

nguyenbahiep1

[laTEX](1+x)^{2013} = C_{2013}^0 + C_{2013}^1.x + C_{2013}^2.x^2 + ..+C_{2013}^{2013}x^{2013} \\ \\ \Leftrightarrow 2013(1+x)^{2012} = C_{2013}^1 + 2C_{2013}^2x+ 3C_{2013}^3.x^2 + ....+ 2013C_{2013}^{2013}x^{2012} \\ \\ \Leftrightarrow 2013(1+x)^{2012}.x = C_{2013}^1x + 2C_{2013}^2x^2+ 3C_{2013}^3.x^3 + ....+ 2013C_{2013}^{2013}x^{2013} \\ \\ \Leftrightarrow 2013(1+x)^{2012} + 2013.2012.x.(1+x)^{2011} = 1^2.C_{2013}^1 + 2^2C_{2013}^2x+ 3^2C_{2013}^3.x^2 + ....+ 2013^2C_{2013}^{2013}x^{2012} \\ \\ x = 1 \Rightarrow 2013.2^{2012} + 2013.2012.2^{2011} = 1^2.C_{2013}^1 + 2^2C_{2013}^2+ 3^2C_{2013}^3 + ....+ 2013^2C_{2013}^{2013}[/laTEX]


Vậy tổng trên bằng

[laTEX]2^{2011}.2013.2014[/laTEX]
 
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