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a fost răspuns

Sa se calculeze
[tex]C _{2011} ^{1} + C_{2011} ^ {2} + C_ {2011} ^ {3} ...+C_ {2011} ^ {2011} =[/tex]

Răspuns :

GreenEyes71
Salut,

[tex](1+1)^{2011}=C_{2011}^0+\underbrace{C_{2011}^1+C_{2011}^2+C_{2011}^3+\ldots+C_{2011}^{2011}}_{S\ =\ Suma\ din\ enun\c{t}},\\\\deci\ S=2^{2011}-C_{2011}^0=2^{2011}-1.\ Simplu,\ nu\ ?[/tex]

Green eyes.

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