Question

Find the coefficient of $\frac{y^3}{x^8}$ in $(x+y)^{-5}$ ?

Answer 1

To find the coefficient of $\frac{y^3}{x^8}\,in\,(x+y)^{-5}$, we can use the binomial theorem. According to the binomial theorem, the expansion of $(x+y)^n$ is given by:
$(x+y)^n$ = $C(n,0)\times x^n\times y^0+C(n,1)\times x^(n-1)\times y^1+C(n,2)\times x^{(n-2)}\times y^2+........+C(n,n-1)\times x^1\times y^{(n-1)}+C(n,n)\times x^0\times y^n$
Where C(n, k) represents the binomial coefficient, given by $C(n,k)=\frac{n!}{(k!\times(n-k)!~)}$
In this case, we have $(x+y)^{-5}$.
Now, let's focus on the term involving y^3/x^8, which corresponds to the term with k = 3 and n - k = 8:
$C(-5,3)\times x^{-4}\times y^3$
The binomial coefficient C(-5, 3) can be calculated as:
$C(-5,3)=\frac{(-5)!}{(3!\times(-5-3)!)}=\frac{(-5)!}{3!\times(-8)!}$
Since factorial values of negative numbers are not defined, C(-5, 3) is not defined. Therefore, the coefficient of $\frac{y^3}{x^8}\,in\,(x+y)^{-5}$ is zero.