Question

Series summation of $S_n$ in closed form?

$\begin{align*} &S_n = \frac{1}{1.2.3.4} + \frac{1}{2.3.4.5} + \frac{1}{3.4.5.6} + \dots + \frac{1}{n.(n+1).(n+2).(n+3)} \end{align*}$

Answer 1

Take your final term as $$\frac{1}{(n-2)(n-1)(n)(n+1)}$$

If we try to write it in another form, we have $$\frac{1}{3}\left ( \frac{1}{(n-2)(n-1)(n)}-\frac{1}{(n-1)(n)(n+1)} \right )$$

OR

$$\frac{1}{3}\left ( X_{n}-X_{n+1}\right )$$

Can you proceed from here by taking summation ?

Comments

Ans: 1/3 * [1/1.2.3  -  1/[(n-1) * n * (n + 1)]  ]  ??

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how it is equal to Xn-Xn+1 did not get it

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That's a general term.