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  • Usuário do Brainly
2014-09-02T22:30:54-03:00
(2x+y)^5

Em geral,

(a+b)^{n}=\dbinom{n}{0}a^{n}b^0+\dbinom{n}{1}a^{n-1}b^1+\dots+\dbinom{n}{n-1}a^{1}b^{n-1}+\dbinom{n}{n}a^0b^{n}

Assim:

(2x+y)^5=\dbinom{5}{0}(2x)^5y^0+\dbinom{5}{1}(2x)^4y^1+\dots+\dbinom{5}{4}(2x)^1y^4+\dbinom{5}{5}(2x)^0y^5

A soma dos coeficientes de (2x+y)^5 é:

\dbinom{5}{0}2^5+\dbinom{5}{1}2^4+\dbinom{5}{2}2^3+\dbinom{5}{3}2^2+\dbinom{5}{4}2^1+\dbinom{5}{5}2^0

=2^5+5\cdot2^4+10\cdot2^3+10\cdot2^2+5\cdot2+1

=32+80+80+40+10+1=243

Alternativa C
4 4 4