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

  • Option 1)

    \frac{55}{336}

  • Option 2)

    \frac{17}{105}

  • Option 3)

    \frac{1}{6}

  • Option 4)

    \frac{19}{112}

 

Answers (1)

As we learnt in 

General term of the sequence -

T_{n}= S_{n}-S_{n-1}

- wherein

We can obtain general term by subtracting S_{n} and S_{n-1}

 

\sum_{n=1}^{5} \: \frac{1}{n\left ( n+1 \right )\left ( n+2 \right )\left ( n+3 \right )}= \frac{k}{3}

Let T_{n}= \frac{1}{n(n+1)(n+2)(n+3)}

T_{n-1}= \frac{1}{n(n+1)(n+2)}

\therefore \: T_{n}-T_{n-1}=-3V_{n}

\therefore \: V_{n}= -\frac{1}{3}\left [ T_{n} -T_{n-1}\right ]

S_{n}= -\frac{1}{3}\left [ T_{n}-T_{0} \right ]

      = -\frac{1}{3}\left [ \frac{1}{(n+1) (n+2) (n+3)} -\frac{1}{6}\right ]

      = \frac{1}{18}-\frac{1}{3\left ( n+1 \right )(n+2)(n+3)}

S_{n}= \frac{n(n^{2}+6n+11)}{18(n+1)(n+2)(n+3)}

Put n=5 we get 

     \frac{5(25+30+11)}{18(6)(7)(8)}= \frac{k}{3}

            = \frac{5\left ( 11 \right )}{6\times 7\times 8}= k

            = \frac{55}{42\times 8}= \frac{55}{336}= k

 


Option 1)

\frac{55}{336}

This option is correct

Option 2)

\frac{17}{105}

This option is incorrect

Option 3)

\frac{1}{6}

This option is incorrect

Option 4)

\frac{19}{112}

This option is incorrect

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