Question

prove: the difference of the reciprocals of two successive integers equals the product of their reciprocals.\\(\frac{1}{n} - \frac{1}{n+1} = \frac{?}{n(n+1)} = \frac{quad}{n} cdot \frac{quad}{n+1}\\)

Explanation:

Step1: Find common denominator

To subtract the fractions, use $n(n+1)$ as the common denominator.
$\frac{1}{n} - \frac{1}{n+1} = \frac{(n+1) - n}{n(n+1)}$

Step2: Simplify numerator

Subtract the terms in the numerator.
$\frac{(n+1) - n}{n(n+1)} = \frac{n+1-n}{n(n+1)} = \frac{1}{n(n+1)}$

Step3: Rewrite as product of reciprocals

Express the fraction as a product of two reciprocals.
$\frac{1}{n(n+1)} = \frac{1}{n} \cdot \frac{1}{n+1}$

Answer:

The missing value in the green box is $1$, so the full proof is:
$\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)} = \frac{1}{n} \cdot \frac{1}{n+1}$