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.