If $$I_1 = \int_0^1 (1-x^{50})^{100}\,dx$$ and $$I_2 = \int_0^1 (1-x^{50})^{101}\,dx$$ such that $$I_2 = \alpha I_1$$ then $$\alpha$$ equals to:

We begin with the two given integrals

$$I_1=\int_0^1\bigl(1-x^{50}\bigr)^{100}\,dx\quad\text{and}\quad I_2=\int_0^1\bigl(1-x^{50}\bigr)^{101}\,dx.$$

To handle both together, it is convenient to define a general integral

$$I_n=\int_0^1\bigl(1-x^{50}\bigr)^{\,n}\,dx,\qquad n\in\mathbb N.$$

We will express $$I_n$$ in Beta-function form. First we perform the substitution

$$t=x^{50}\quad\Longrightarrow\quad x=t^{1/50},\qquad dx=\frac1{50}\,t^{\frac1{50}-1}\,dt.$$

Under this substitution, when $$x=0$$ we have $$t=0$$, and when $$x=1$$ we have $$t=1$$. So the limits remain $$0$$ to $$1$$. Substituting into the integrand we obtain

$$I_n = \int_0^1\bigl(1-t\bigr)^{\,n}\;\frac1{50}\,t^{\frac1{50}-1}\,dt.$$

Re-ordering the factors gives

$$I_n = \frac1{50}\int_0^1 t^{\frac1{50}-1}\,(1-t)^{\,n}\,dt.$$

By definition, the Beta function is

$$B(p,q)=\int_0^1 t^{p-1}(1-t)^{q-1}\,dt.$$

Comparing, we identify

$$p=\frac1{50},\qquad q=n+1.$$

Hence

$$I_n = \frac1{50}\,B\!\Bigl(\tfrac1{50},\,n+1\Bigr).$$

Now we recall the relation between the Beta and Gamma functions:

$$B(p,q)=\frac{\Gamma(p)\,\Gamma(q)}{\Gamma(p+q)}.$$

Applying this formula we get

$$I_n=\frac1{50}\; \frac{\Gamma\!\bigl(\tfrac1{50}\bigr)\, \Gamma(n+1)} {\Gamma\!\bigl(n+1+\tfrac1{50}\bigr)}.$$

We only need the ratio $$\alpha=\frac{I_2}{I_1}=\frac{I_{101}}{I_{100}}$$ because $$I_2=I_{101}$$ and $$I_1=I_{100}$$.

Using the Gamma-function expression,

$$ \frac{I_{101}}{I_{100}} =\frac{\dfrac1{50}\, \Gamma\!\bigl(\tfrac1{50}\bigr)\, \Gamma(102)\big/ \Gamma\!\bigl(102+\tfrac1{50}\bigr)} {\dfrac1{50}\, \Gamma\!\bigl(\tfrac1{50}\bigr)\, \Gamma(101)\big/ \Gamma\!\bigl(101+\tfrac1{50}\bigr)}.$$

The constant factors $$\dfrac1{50}$$ and $$\Gamma\!\bigl(\tfrac1{50}\bigr)$$ cancel, so

$$ \frac{I_{101}}{I_{100}} =\frac{\Gamma(102)\,\Gamma\!\bigl(101+\tfrac1{50}\bigr)} {\Gamma(101)\,\Gamma\!\bigl(102+\tfrac1{50}\bigr)}. $$

We next employ the fundamental Gamma-function identity

$$\Gamma(z+1)=z\,\Gamma(z).$$

Applying it to both $$\Gamma(102)$$ and $$\Gamma\!\bigl(102+\tfrac1{50}\bigr)$$:

$$\Gamma(102) = 101\,\Gamma(101),$$

$$\Gamma\!\Bigl(102+\tfrac1{50}\Bigr) =\Bigl(101+\tfrac1{50}\Bigr)\, \Gamma\!\Bigl(101+\tfrac1{50}\Bigr).$$

Substituting these back, we obtain

$$ \frac{I_{101}}{I_{100}} =\frac{101\,\Gamma(101)\, \Gamma\!\bigl(101+\tfrac1{50}\bigr)} {\Gamma(101)\, \bigl(101+\tfrac1{50}\bigr)\, \Gamma\!\bigl(101+\tfrac1{50}\bigr)}. $$

Both $$\Gamma(101)$$ and $$\Gamma\!\bigl(101+\tfrac1{50}\bigr)$$ cancel, leaving

$$\frac{I_{101}}{I_{100}}=\frac{101}{\,101+\tfrac1{50}}.$$

We now convert the denominator to a single fraction:

$$101+\tfrac1{50} =\frac{101\cdot50}{50}+\frac1{50} =\frac{5050+1}{50} =\frac{5051}{50}.$$

Hence

$$\frac{I_{101}}{I_{100}} =\frac{101}{\,5051/50} =101\cdot\frac{50}{5051} =\frac{5050}{5051}.$$

But by definition of $$\alpha$$ we have $$I_2=\alpha I_1$$, so

$$\alpha=\frac{I_2}{I_1}=\frac{5050}{5051}.$$

Among the given choices, this value matches Option C.

Hence, the correct answer is Option C.