Question

Logan has to find the sum of the given fractions. Being the smart boy he is, Logan splits each fraction into two so that he does not need to add the fractions directly. Follow Logan's method and find the sum:
$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{4}$

• A. $\frac{3}{4}$
• B. $1$
• C. $\frac{4}{5}$
• D. $2$

Answer 1

The correct option is B $1$

Each term, in this sum can be expressed as the difference between two fractions as:

$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{4}$

$=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\frac{1}{4}$

Grouping the fraction together so that they cancel out, we get:

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

$=1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+(-\frac{1}{4}+\frac{1}{4})$

$=1$