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Question: simplify $$\frac{(k-1)!}{(k+2)!}$$

What I did was:

$$\frac{(k - 1)!k!}{(k + 2)! \cdot (k + 1)!}$$ This I did following the rule $n! = n \times (n - 1)!$.

can this be simplified further? Thanks.


marked as duplicate by davidlowryduda Jul 24 '16 at 8:26

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  • 3
    $\begingroup$ @Utsav Did you not learn anything e.g. from answer and comments of Zain on your former question: "The general strategy with factorials is to manipulate the bigger number to display some terms with a factor of the smaller number"? Your question is very very close to your former one and the same principles can be applied. $\endgroup$ – drhab Jul 23 '16 at 10:47
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    $\begingroup$ Almost a duplicate. If the OP would have taken good notice of the answers to his former question then asking this would have been unnecessary. $\endgroup$ – drhab Jul 23 '16 at 10:57
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    $\begingroup$ We must be careful with judgements and this is true for you as well. Btw, welcome at SE, and do not get discouraged by this. The site is meant for training mathematical skills. If two sortlike questions arrive within short time then it looks as if the OP is only after answers and not the training of his skills. We must discourage that behaviour, but sometimes step over the line. I don't believe that you judge me as a fool. Again: welcome and enjoy this site, youngster! $\endgroup$ – drhab Jul 23 '16 at 11:07
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    $\begingroup$ @Utsav Please observe that $(k+2)! \neq (k + 2)! \times (k + 1)!$, rather $(k+2)! =(k + 2) \times (k + 1)!$. $\endgroup$ – Olivier Oloa Jul 23 '16 at 11:09
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    $\begingroup$ Hello --- you have requested that this question be deleted. This would not be fair to those kind users who have taken the time to answer your question, or to those who have liked the answers and upvoted them. We very rarely delete upvoted content, and the system prohibits this for a reason. Good luck with your math studies. $\endgroup$ – davidlowryduda Jul 24 '16 at 8:28

One may write $$ \frac{(k-1)!}{(k+2)!}=\frac{1\cdot\color{blue}{(k-1)!}}{(k+2)(k+1)k\:\color{blue}{(k-1)!}}=\frac1{(k+2)(k+1)k}. $$

  • $\begingroup$ The OP question above is fine to me. Let's try not to overlook too much some questions on a basis of level. No problem @Behrouz Maleki $\endgroup$ – Olivier Oloa Jul 23 '16 at 11:04

HInt: $(k+2)!=(k+2)(k+1)k(k-1)!$


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