If _n=1^5(1)/(n(n+1)(n+2)(n+3))=(k)/(3), then k is equal to:

Question

If $\displaystyle\sum_{n=1}^{5}\frac{1}{n(n+1)(n+2)(n+3)}=\frac{k}{3}$, then $k$ is equal to:

Accepted Answer

Using the telescoping identity
$$\frac{1}{n(n+1)(n+2)(n+3)}=\frac13\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right],$$
the sum for $n=1$ to $5$ telescopes to
$$\frac13\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{6\cdot7\cdot8}\right]=\frac13\left[\frac16-\frac{1}{336}\right]=\frac13\times\frac{55}{336}=\frac{55}{1008}=\frac{k}{3}.$$

$$k=\frac{55}{336}.$$

Solutions · 1