As @Jennifer answered, the general term seems to be \frac{\prod_{i=0}^{n-1} (5i+1) }{\prod_{i=0}^{n-1} 5(i+1)}=\frac{\Gamma\left(\frac15+n\right)}{\Gamma\left(\frac15\right)\Gamma\left(n+1\right)}=\frac{\left(\frac15\right)_n}{n!} \tag1 ...

E\left(\left(1-\frac{a}{n}\right)^X\right) = \sum_{x=0}^{\infty} \left(1-\frac{a}{n}\right)^x e^{-\lambda}\frac{\lambda^x}{x!} = e^{-\lambda} e^{\lambda\left(1-\frac{a}{n}\right)} = e^{\frac{-a\lambda}{n}} ...

In these formulas, we hav that X = (x(t), y(t), z(t)). Therefore, we begin by calculating the derivatives of the vectors component-wise with respect to t and have X' = (x'(t), y'(t),z'(t)) = \left( e^t \sin(t) + e^t \cos(t) , e^t \cos(t) - e^t \sin(t), e^t\right) ...

(A) C is closed because the elements of C are those invertible matrices (a_{ij})_{1\leqslant i,j\leqslant2n} such that a_{ij}=0 when i>n and j\leqslant n.

You want to find all x\in{\mathbb C} such that the set (x+3)^{1/3}\cap (x-1)^{1/2}\subset{\mathbb C} is nonempty. In other words, you want to find all x\in{\mathbb C} such that there is ...

Your idea is good, but it's not laid out well. An identity can be verified by transforming one side into the other or both into the same expression. In this case it's easier starting from the ...