Evaluate the following expression:

[tex]\[ f\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x+1}\right)\left(1+\frac{1}{x+2}\right)\left(1+\frac{1}{x+3}\right) \ldots \left(1+\frac{1}{x+n}\right) \][/tex]



Answer :

To solve the problem of simplifying the product \(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x+1}\right)\left(1+\frac{1}{x+2}\right) \cdots \left(1+\frac{1}{x+n}\right)\), let's go through it step by step.

We need to simplify the expression:

[tex]\[ P = \prod_{i=0}^{n} \left(1 + \frac{1}{x+i}\right) \][/tex]

Let's consider each individual term inside the product:

[tex]\[ \left(1 + \frac{1}{x+i}\right) \][/tex]

This term can be rewritten as:

[tex]\[ 1 + \frac{1}{x+i} = \frac{(x+i)+1}{x+i} = \frac{x+i+1}{x+i} \][/tex]

So for each \(i\), we have:

[tex]\[ \left(1 + \frac{1}{x+i}\right) = \frac{x+i+1}{x+i} \][/tex]

Now, substituting this back into the product, we have:

[tex]\[ P = \prod_{i=0}^{n} \frac{x+i+1}{x+i} \][/tex]

This product is a telescoping product. In a telescoping product, most of the terms cancel out each other. Let’s write out a few terms to see this cancellation:

[tex]\[ P = \frac{x+1}{x} \cdot \frac{x+2}{x+1} \cdot \frac{x+3}{x+2} \cdot \ldots \cdot \frac{x+n+1}{x+n} \][/tex]

Notice that in this product:
- The term \(x+1\) in the numerator of the first fraction cancels with the denominator of the second fraction.
- The term \(x+2\) in the numerator of the second fraction cancels with the denominator of the third fraction.
- This pattern continues until the last term, where \(x+n\) in the numerator cancels with \(x+n\) in the denominator of the last fraction.

All intermediate terms cancel out, leaving us with:

[tex]\[ P = \frac{x+n+1}{x} \][/tex]

Therefore, the simplified form of the given product is:

[tex]\[ \boxed{\frac{x+n+1}{x}} \][/tex]