Is there any way to simplify the following fractions? Thank you!












0












$begingroup$


Is there any way to simplify the following fractions?



$[n^{3}-1]/[n^{4}-1]$



$n!/n^{n}$






fractions







share|cite|improve this question









$endgroup$












  • $begingroup$
    You can factor an $n-1$ from both numerator and denominator in your first expression.
    $endgroup$
    – coreyman317
    Jan 24 at 3:40










  • $begingroup$
    As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
    $endgroup$
    – coreyman317
    Jan 24 at 3:43


















0












$begingroup$


Is there any way to simplify the following fractions?



$[n^{3}-1]/[n^{4}-1]$



$n!/n^{n}$






fractions







share|cite|improve this question









$endgroup$












  • $begingroup$
    You can factor an $n-1$ from both numerator and denominator in your first expression.
    $endgroup$
    – coreyman317
    Jan 24 at 3:40










  • $begingroup$
    As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
    $endgroup$
    – coreyman317
    Jan 24 at 3:43
















0












0








0





$begingroup$


Is there any way to simplify the following fractions?



$[n^{3}-1]/[n^{4}-1]$



$n!/n^{n}$






fractions







share|cite|improve this question









$endgroup$




Is there any way to simplify the following fractions?



$[n^{3}-1]/[n^{4}-1]$



$n!/n^{n}$






fractions




fractions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 3:37









JennyJenny

624




624












  • $begingroup$
    You can factor an $n-1$ from both numerator and denominator in your first expression.
    $endgroup$
    – coreyman317
    Jan 24 at 3:40










  • $begingroup$
    As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
    $endgroup$
    – coreyman317
    Jan 24 at 3:43




















  • $begingroup$
    You can factor an $n-1$ from both numerator and denominator in your first expression.
    $endgroup$
    – coreyman317
    Jan 24 at 3:40










  • $begingroup$
    As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
    $endgroup$
    – coreyman317
    Jan 24 at 3:43


















$begingroup$
You can factor an $n-1$ from both numerator and denominator in your first expression.
$endgroup$
– coreyman317
Jan 24 at 3:40




$begingroup$
You can factor an $n-1$ from both numerator and denominator in your first expression.
$endgroup$
– coreyman317
Jan 24 at 3:40












$begingroup$
As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
$endgroup$
– coreyman317
Jan 24 at 3:43






$begingroup$
As for the second, pick a random integer and write out the numerator and denominator as $frac{1cdot 2 cdot...cdot n}{n cdot n cdot...cdot n}$ and see which number can cancel.
$endgroup$
– coreyman317
Jan 24 at 3:43












2 Answers
2






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oldest

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0












$begingroup$


  1. Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.


  2. Not much information is given except the common $n$.







share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    For the first:



    $$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$



    For the second:
    $$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
      $endgroup$
      – Rhys Hughes
      Jan 24 at 4:17











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$


    1. Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.


    2. Not much information is given except the common $n$.







    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$


      1. Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.


      2. Not much information is given except the common $n$.







      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$


        1. Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.


        2. Not much information is given except the common $n$.







        share|cite|improve this answer









        $endgroup$




        1. Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.


        2. Not much information is given except the common $n$.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 24 at 3:48









        MonkeyKingMonkeyKing

        2,3211029




        2,3211029























            0












            $begingroup$

            For the first:



            $$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$



            For the second:
            $$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
              $endgroup$
              – Rhys Hughes
              Jan 24 at 4:17
















            0












            $begingroup$

            For the first:



            $$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$



            For the second:
            $$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
              $endgroup$
              – Rhys Hughes
              Jan 24 at 4:17














            0












            0








            0





            $begingroup$

            For the first:



            $$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$



            For the second:
            $$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$






            share|cite|improve this answer









            $endgroup$



            For the first:



            $$frac{n^3-1}{n^4-1}=frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-frac{n^3}{1+n+n^2+n^3}$$



            For the second:
            $$frac{n!}{n^n}=frac{(n-1)!}{n^{n-1}}=frac{(1-frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=frac{(n-k-1)!prod_{r=1}^{k}(1-frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k in Bbb N$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 24 at 4:12









            Rhys HughesRhys Hughes

            6,9571530




            6,9571530












            • $begingroup$
              If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
              $endgroup$
              – Rhys Hughes
              Jan 24 at 4:17


















            • $begingroup$
              If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
              $endgroup$
              – Rhys Hughes
              Jan 24 at 4:17
















            $begingroup$
            If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
            $endgroup$
            – Rhys Hughes
            Jan 24 at 4:17




            $begingroup$
            If you're wondering how I got this, note that $$n(1-frac 1n)=n-1$$
            $endgroup$
            – Rhys Hughes
            Jan 24 at 4:17


















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