Question

Find the value of $\frac{a}{2}+\frac{a^2}{4}+\frac{a^3}{8}-\frac{1}{6}$.

• A. $\frac{12a+6a^2+a^3-4}{24}$
• B. $\frac{12a+6a^2+3a^3-4}{12}$
• C. $\frac{12a+6a^2+3a^3-24}{24}$
• D. $\frac{12a+6a^2+3a^3-4}{24}$

Answer 1

The correct option is D $\frac{12a+6a^2+3a^3-4}{24}$

$\frac{a}{2}+\frac{a^2}{4}+\frac{a^3}{8}-\frac{1}{6}$
First find the LCM of denominator, $2,4,8$ and $6$
So, the LCM of $2,4,6$ and 8 is $24$
Therefore, $\frac{12a+6a^2+3a^3-4}{24}$