1. We need to evaluate

$$I=\int_0^1 (1-x^{10})^{20}\,dx$$

in the form

$$I=a\times \beta(b,c),$$

where

$$\beta(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,dx.$$

2. Use the substitution

$$t=x^{10} \quad \Rightarrow \quad x=t^{1/10},$$

so that

$$dx=\frac{1}{10}t^{-9/10}\,dt.$$

Also, when $x=0$, $t=0$, and when $x=1$, $t=1$.

3. Substitute into the integral:

$$I=\int_0^1 (1-t)^{20}\cdot \frac{1}{10}t^{-9/10}\,dt.$$

That is,

$$I=\frac{1}{10}\int_0^1 t^{-9/10}(1-t)^{20}\,dt.$$

4. Compare this with the Beta-function form

$$\beta(m,n)=\int_0^1 t^{m-1}(1-t)^{n-1}\,dt.$$

We identify:

• $m-1=-\frac{9}{10}$, so $m=\frac{1}{10};$
• $n-1=20$, so $n=21.$

Hence,

$$I=\frac{1}{10}\beta\left(\frac{1}{10},21\right).$$

Therefore,

$$a=\frac{1}{10},\quad b=\frac{1}{10},\quad c=21.$$

5. Now compute

$$100(a+b+c)=100\left(\frac{1}{10}+\frac{1}{10}+21\right).$$

Since

$$\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5},$$

we get

$$100\left(21+\frac{1}{5}\right)=100\cdot \frac{106}{5}=20\cdot 106=2120.$$

6. Therefore the correct option is

$$\boxed{\text{D: }2120}.$$

7. Comparison with stored answer: The stored correct answer is D, which matches our result.