# If then k is equal to: Option 1) Option 2) Option 3) Option 4)

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)

This option is correct

Option 2)

This option is incorrect

Option 3)

This option is incorrect

Option 4)

This option is incorrect

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