Simplifying the Algebraic Product

The problem requires simplifying the product of four terms:

$ (1 + \frac{1}{x})(1 + \frac{1}{x+1})(1 + \frac{1}{x+2})(1 + \frac{1}{x+3}) $

Step-by-Step Simplification

First, rewrite each term by finding a common denominator:

• $1 + \frac{1}{x} = \frac{x}{x} + \frac{1}{x} = \frac{x+1}{x}$
• $1 + \frac{1}{x+1} = \frac{x+1}{x+1} + \frac{1}{x+1} = \frac{x+2}{x+1}$
• $1 + \frac{1}{x+2} = \frac{x+2}{x+2} + \frac{1}{x+2} = \frac{x+3}{x+2}$
• $1 + \frac{1}{x+3} = \frac{x+3}{x+3} + \frac{1}{x+3} = \frac{x+4}{x+3}$

Now, substitute these back into the original product:

$ (\frac{x+1}{x}) \times (\frac{x+2}{x+1}) \times (\frac{x+3}{x+2}) \times (\frac{x+4}{x+3}) $

Applying Cancellation (Telescoping Product)

Observe that terms cancel out diagonally:

• The numerator $(x+1)$ of the first term cancels with the denominator $(x+1)$ of the second term.
• The numerator $(x+2)$ of the second term cancels with the denominator $(x+2)$ of the third term.
• The numerator $(x+3)$ of the third term cancels with the denominator $(x+3)$ of the fourth term.

After cancellation, the expression simplifies to:

$ \frac{\cancel{x+1}}{x} \times \frac{\cancel{x+2}}{\cancel{x+1}} \times \frac{\cancel{x+3}}{\cancel{x+2}} \times \frac{x+4}{\cancel{x+3}} = \frac{x+4}{x} $

The simplified expression is $\frac{x+4}{x}$.