Sigma laws problem / sigma distribution












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]1



Hi
Can somebody explain to me how that was made using sigma laws .
Thank you for help in advance






calculus probability







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    0












    $begingroup$


    ]1



    Hi
    Can somebody explain to me how that was made using sigma laws .
    Thank you for help in advance






    calculus probability







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      ]1



      Hi
      Can somebody explain to me how that was made using sigma laws .
      Thank you for help in advance






      calculus probability







      share|cite|improve this question









      $endgroup$




      ]1



      Hi
      Can somebody explain to me how that was made using sigma laws .
      Thank you for help in advance






      calculus probability




      calculus probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 19:08









      Philip MichalowskiPhilip Michalowski

      61




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          1 Answer
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          $begingroup$

          Well this is easy! Rewrite the sum as
          $$sum_{k=1}^{n}n-k=sum_{k=1}^n n-sum_{k=1}^{n}k=n^2-frac{n(n+1)}{2}=n^2-frac{n^2(1+tfrac{1}{n})}{2}=n^2Big(frac{2-1-tfrac{1}{n}}{2}Big)=n^2Big(frac{1-tfrac{1}{n}}{2}Big)$$
          No sigma laws were used here!



          Edit:
          $$sum_{k=1}^n n=underbrace{n+cdots+n}_{ntext{ times}}=n^2$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ok so how do you solve the first sum ∑n . Do you just multiply by n ?
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:25










          • $begingroup$
            $$sum_{k=1}^n n=nsum_{k=1}^n 1=nunderbrace{(1+cdots+1)}_{text{n times}}=n^2$$
            $endgroup$
            – John Doe
            Jan 7 at 19:33












          • $begingroup$
            @PhilipMichalowski $$sumlimits_{k=1}^n n=underbrace{n+n+...+n}_{textrm{n-times}}=ncdot n$$
            $endgroup$
            – callculus
            Jan 7 at 19:34












          • $begingroup$
            @JohnDoe oh ok , thank you very much
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:34











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          Well this is easy! Rewrite the sum as
          $$sum_{k=1}^{n}n-k=sum_{k=1}^n n-sum_{k=1}^{n}k=n^2-frac{n(n+1)}{2}=n^2-frac{n^2(1+tfrac{1}{n})}{2}=n^2Big(frac{2-1-tfrac{1}{n}}{2}Big)=n^2Big(frac{1-tfrac{1}{n}}{2}Big)$$
          No sigma laws were used here!



          Edit:
          $$sum_{k=1}^n n=underbrace{n+cdots+n}_{ntext{ times}}=n^2$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ok so how do you solve the first sum ∑n . Do you just multiply by n ?
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:25










          • $begingroup$
            $$sum_{k=1}^n n=nsum_{k=1}^n 1=nunderbrace{(1+cdots+1)}_{text{n times}}=n^2$$
            $endgroup$
            – John Doe
            Jan 7 at 19:33












          • $begingroup$
            @PhilipMichalowski $$sumlimits_{k=1}^n n=underbrace{n+n+...+n}_{textrm{n-times}}=ncdot n$$
            $endgroup$
            – callculus
            Jan 7 at 19:34












          • $begingroup$
            @JohnDoe oh ok , thank you very much
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:34
















          1












          $begingroup$

          Well this is easy! Rewrite the sum as
          $$sum_{k=1}^{n}n-k=sum_{k=1}^n n-sum_{k=1}^{n}k=n^2-frac{n(n+1)}{2}=n^2-frac{n^2(1+tfrac{1}{n})}{2}=n^2Big(frac{2-1-tfrac{1}{n}}{2}Big)=n^2Big(frac{1-tfrac{1}{n}}{2}Big)$$
          No sigma laws were used here!



          Edit:
          $$sum_{k=1}^n n=underbrace{n+cdots+n}_{ntext{ times}}=n^2$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ok so how do you solve the first sum ∑n . Do you just multiply by n ?
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:25










          • $begingroup$
            $$sum_{k=1}^n n=nsum_{k=1}^n 1=nunderbrace{(1+cdots+1)}_{text{n times}}=n^2$$
            $endgroup$
            – John Doe
            Jan 7 at 19:33












          • $begingroup$
            @PhilipMichalowski $$sumlimits_{k=1}^n n=underbrace{n+n+...+n}_{textrm{n-times}}=ncdot n$$
            $endgroup$
            – callculus
            Jan 7 at 19:34












          • $begingroup$
            @JohnDoe oh ok , thank you very much
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:34














          1












          1








          1





          $begingroup$

          Well this is easy! Rewrite the sum as
          $$sum_{k=1}^{n}n-k=sum_{k=1}^n n-sum_{k=1}^{n}k=n^2-frac{n(n+1)}{2}=n^2-frac{n^2(1+tfrac{1}{n})}{2}=n^2Big(frac{2-1-tfrac{1}{n}}{2}Big)=n^2Big(frac{1-tfrac{1}{n}}{2}Big)$$
          No sigma laws were used here!



          Edit:
          $$sum_{k=1}^n n=underbrace{n+cdots+n}_{ntext{ times}}=n^2$$






          share|cite|improve this answer











          $endgroup$



          Well this is easy! Rewrite the sum as
          $$sum_{k=1}^{n}n-k=sum_{k=1}^n n-sum_{k=1}^{n}k=n^2-frac{n(n+1)}{2}=n^2-frac{n^2(1+tfrac{1}{n})}{2}=n^2Big(frac{2-1-tfrac{1}{n}}{2}Big)=n^2Big(frac{1-tfrac{1}{n}}{2}Big)$$
          No sigma laws were used here!



          Edit:
          $$sum_{k=1}^n n=underbrace{n+cdots+n}_{ntext{ times}}=n^2$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 7 at 19:34

























          answered Jan 7 at 19:20









          FakemistakeFakemistake

          1,692815




          1,692815












          • $begingroup$
            ok so how do you solve the first sum ∑n . Do you just multiply by n ?
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:25










          • $begingroup$
            $$sum_{k=1}^n n=nsum_{k=1}^n 1=nunderbrace{(1+cdots+1)}_{text{n times}}=n^2$$
            $endgroup$
            – John Doe
            Jan 7 at 19:33












          • $begingroup$
            @PhilipMichalowski $$sumlimits_{k=1}^n n=underbrace{n+n+...+n}_{textrm{n-times}}=ncdot n$$
            $endgroup$
            – callculus
            Jan 7 at 19:34












          • $begingroup$
            @JohnDoe oh ok , thank you very much
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:34


















          • $begingroup$
            ok so how do you solve the first sum ∑n . Do you just multiply by n ?
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:25










          • $begingroup$
            $$sum_{k=1}^n n=nsum_{k=1}^n 1=nunderbrace{(1+cdots+1)}_{text{n times}}=n^2$$
            $endgroup$
            – John Doe
            Jan 7 at 19:33












          • $begingroup$
            @PhilipMichalowski $$sumlimits_{k=1}^n n=underbrace{n+n+...+n}_{textrm{n-times}}=ncdot n$$
            $endgroup$
            – callculus
            Jan 7 at 19:34












          • $begingroup$
            @JohnDoe oh ok , thank you very much
            $endgroup$
            – Philip Michalowski
            Jan 7 at 19:34
















          $begingroup$
          ok so how do you solve the first sum ∑n . Do you just multiply by n ?
          $endgroup$
          – Philip Michalowski
          Jan 7 at 19:25




          $begingroup$
          ok so how do you solve the first sum ∑n . Do you just multiply by n ?
          $endgroup$
          – Philip Michalowski
          Jan 7 at 19:25












          $begingroup$
          $$sum_{k=1}^n n=nsum_{k=1}^n 1=nunderbrace{(1+cdots+1)}_{text{n times}}=n^2$$
          $endgroup$
          – John Doe
          Jan 7 at 19:33






          $begingroup$
          $$sum_{k=1}^n n=nsum_{k=1}^n 1=nunderbrace{(1+cdots+1)}_{text{n times}}=n^2$$
          $endgroup$
          – John Doe
          Jan 7 at 19:33














          $begingroup$
          @PhilipMichalowski $$sumlimits_{k=1}^n n=underbrace{n+n+...+n}_{textrm{n-times}}=ncdot n$$
          $endgroup$
          – callculus
          Jan 7 at 19:34






          $begingroup$
          @PhilipMichalowski $$sumlimits_{k=1}^n n=underbrace{n+n+...+n}_{textrm{n-times}}=ncdot n$$
          $endgroup$
          – callculus
          Jan 7 at 19:34














          $begingroup$
          @JohnDoe oh ok , thank you very much
          $endgroup$
          – Philip Michalowski
          Jan 7 at 19:34




          $begingroup$
          @JohnDoe oh ok , thank you very much
          $endgroup$
          – Philip Michalowski
          Jan 7 at 19:34


















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