Answer to Question #140389 in Discrete Mathematics for Promise Omiponle

Question #140389
What is the coefficient of x^3 y^13 in the expansion of (−1x+2y)^16?
1
Expert's answer
2020-11-09T15:30:38-0500

By Binomial Theorem


(a+b)n=(n0)anb0+(n1)an1b1+...+(a+b)^n=\dbinom{n}{0}a^nb^0+\dbinom{n}{1}a^{n-1}b^1+...+

+(nk)akbnk+...+(nn)a0bn+\dbinom{n}{k}a^kb^{n-k}+...+\dbinom{n}{n}a^0b^n

We have a=x,b=2y,n=16.a=-x, b=2y, n=16.

We find the coefficient of x3y13x^3 y^{13}


k=3,(nk)akbnk=(163)(x)3(2y)163=k=3, \dbinom{n}{k}a^kb^{n-k}=\dbinom{16}{3}(-x)^3(2y)^{16-3}=

=16!3!(163)!(8192)x3y13=560(8192)x3y13==-\dfrac{16!}{3!(16-3)!}(8192)x^3y^{13}=-560(8192)x^3y^{13}=

=4587520x3y13=-4587520x^3y^{13}

The coefficient of x3y3x^3y^{3} in the expansion of (1x+2y)16(-1x+2y)^{16} is 4587520.-4587520.



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