Sequences & Series
Series sum using cube roots of unity
MJMT_Full_Test_01
Grade 12
Question:
The sum $\displaystyle\sum_{n=1}^{\infty}\frac{9n^2}{(3n)!}$ is equal to
$e+\dfrac{1}{\sqrt{e}}\cos\!\left(\dfrac{\sqrt{3}}{2}\right)$
$\dfrac{2}{3}e-\dfrac{1}{\sqrt{e}}\cos\!\left(\dfrac{\sqrt{3}}{2}\right)$
$\dfrac{2}{3}\!\left(e+\dfrac{1}{\sqrt{e}}\cos\!\left(\dfrac{\sqrt{3}}{2}\right)\right)$
$\dfrac{2}{3}\!\left(e-\dfrac{1}{\sqrt{e}}\cos\!\left(\dfrac{\sqrt{3}}{2}\right)\right)$
Step-by-Step Solution
Key Concept: Use the fact that $\sum_{n=0}^\infty x^{3n}/(3n)! = (e^x+e^{\omega x}+e^{\omega^2 x})/3$ where $\omega=e^{2\pi i/3}$. Differentiate to extract coefficients of $n^2/(3n)!$ series.
$\sum = \dfrac{2}{3}\!\left(e-\dfrac{1}{\sqrt{e}}\cos\!\left(\dfrac{\sqrt{3}}{2}\right)\right)$.
Correct Answer: 4