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 $\frac{55}{336}$

${\sum }_{n=1}^5\frac{1}{n(n+1)(n+2)(n+3)}=\frac{k}{3}\frac{1}{2}{\sum }_{n=1}^5(\frac{1}{n(n+3)}-\frac{1}{(n+1)(n+3)})=\frac{k}{3}{\sum }_{n=1}^5[\frac{1}{3}\left(\frac{3}{n(n+3)}\right)-\frac{1}{(n+1)(n+3)}]=\frac{2k}{3}{\sum }_{n=1}^5(\frac{1}{3n}-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{3(n+3)})=\frac{2k}{3}\frac{1}{3}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})-(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6})+(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})-\frac{1}{3}(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})=\frac{2k}{3}\frac{1}{3}(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8})-(\frac{1}{2}-\frac{1}{7})=\frac{2k}{3}\frac{1}{3}\left(\frac{235}{3\times 56}\right)-\frac{5}{14}=\frac{2k}{3}\frac{235}{3\times 56}-\frac{5}{14}=2k\frac{55}{3\times 56}=2kk=\frac{55}{336}$