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]
