Permutations And Combinations Question 326

Question: If $ S=(1)(1!)+(2)(2!)+(3)(3!)+…+n(n!), $ then

Options:

A) $ \frac{S+1}{n!}\in $ integer

B) $ \frac{S+1}{n!}\notin $ Integer

C) $ \frac{S+1}{n!} $ cannot be discussed

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

  • [a] We have, $ S=\sum\limits _{k=1}^{n}{k(k!)}=\sum\limits _{k=1}^{n}{{(k+1)-1}}(k!) $ $ =\sum\limits _{k=1}^{n}{{(k+1)!-k!}=(n+1)!-1\Rightarrow S+1=(n+1)!} $ Thus, $ \frac{S+1}{n!}\in $ integer.

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