The product of $$\left(1 - \frac{1}{x+1}\right) \left(1 - \frac{1}{x+2}\right) \left(1 - \frac{1}{x+3}\right) \cdots \left(1 - \frac{1}{2x}\right)$$​ is:

Correct option is B

​Given:

​$$P(x) = \left(1 - \frac{1}{x+1}\right) \left(1 - \frac{1}{x+2}\right) \left(1 - \frac{1}{x+3}\right) \cdots \left(1 - \frac{1}{2x}\right)$$ 

Solution:

Simplifying each factor of the expression;

Each term in the product has the form $$1 -\frac{1}{ x + k}$$ ​ , where k is an integer ranging from 1 to x. We can rewrite each term as:

​$$\left(1 - \frac{1}{x+k}\right) = \frac{x+k-1}{x+k}$$ 

Therefore, the product becomes:

​$$P(x) = \prod_{k=1}^{x} \frac{x+k-1}{x+k}$$ 

We can write:

​$$P(x) = \frac{x}{x+1} \times \frac{x+1}{x+2} \times \frac{x+2}{x+3} \times \cdots \times \frac{2x-1}{2x}$$ 

Notice that the numerator of each term cancels with the denominator of the next term. So, all intermediate terms cancel out, and we are left with:

​$$P(x) = \frac{x}{2x}$$ 

​$$P(x) = \frac{1}{2}$$ 

thus, The product is:  $$\frac{1}{2}$$ .​