Question

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

• A. $\frac{55}{336}$
• B. $\frac{17}{105}$
• C. $\frac{19}{112}$
• D. $\frac{1}{6}$

Answer 1

The correct option is A $\frac{55}{336}$

${\sum }_{n=1}^5\frac{1}{n(n+1)(n+2)(n+3)}=\frac{k}{3}$
$\frac{1}{\mathrm{1.2.3.4}}+\frac{1}{\mathrm{2.3.4.5}}+\frac{1}{\mathrm{3.4.5.6}}+\frac{1}{\mathrm{4.5.6.7}}+\frac{1}{\mathrm{5.6.7.8}}=\frac{k}{3}$
$\frac{1}{24}+\frac{1}{120}+\frac{1}{360}+\frac{1}{840}+\frac{1}{1680}=\frac{k}{3}$
$\frac{275}{5040}=\frac{k}{3}$
$\Rightarrow k=\frac{55}{336}$