Sequences & Series
Telescoping series — formula identification
MJAT_TS6_P1
Grade 12
Question:
If $\displaystyle\frac{1}{1\cdot 2\cdot 4}+\frac{2}{2\cdot 3\cdot 5}+\frac{3}{3\cdot 4\cdot 6}+\cdots+\frac{n}{n(n+1)(n+3)} = \frac{1}{6}\left[\frac{a}{6}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right]$, where $a,b\in\mathbb{Z}^+$. Then $\dfrac{a}{12}+b$ equals:
Step-by-Step Solution
Key Concept: Partial fraction decomposition: $\frac{n}{n(n+1)(n+3)}=\frac{1}{(n+1)(n+3)}$. Then use: $\frac{1}{(n+1)(n+3)}=\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)$. Telescope to get the closed form.
$a/12+b=\mathbf{2.50}$.
Correct Answer: 2.50