Further explanation for steps of an equation that proofs that $sum^{n}_{k=0}kcdot binom{n}{k}=ncdot2^{n-1}$












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So, this is one of the questions in my textbook, which seems to be quite common: $$sum^{n}_{k=0}kcdot binom{n}{k}=ncdot2^{n-1}$$



The same book provides the following solution:
$$sum_{k=0}^{n}kcdot binom{n}{k}= sum^{n}_{k=0}ncdotbinom{n-1}{k-1}=nsum^{n-1}_{k=0}binom{n-1}{k}=ncdot2^{n-1}$$



What is unclear to me, is (1) how I get from $$sum^{n}_{k=0}ncdotbinom{n-1}{k-1} $$ to$$nsum^{n-1}_{k=0}binom{n-1}{k}$$ and (2) from $$nsum^{n-1}_{k=0}binom{n-1}{k}$$ to $$ncdot2^{n-1}$$ respectively.






sequences-and-series discrete-mathematics summation factorial







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    1












    $begingroup$


    So, this is one of the questions in my textbook, which seems to be quite common: $$sum^{n}_{k=0}kcdot binom{n}{k}=ncdot2^{n-1}$$



    The same book provides the following solution:
    $$sum_{k=0}^{n}kcdot binom{n}{k}= sum^{n}_{k=0}ncdotbinom{n-1}{k-1}=nsum^{n-1}_{k=0}binom{n-1}{k}=ncdot2^{n-1}$$



    What is unclear to me, is (1) how I get from $$sum^{n}_{k=0}ncdotbinom{n-1}{k-1} $$ to$$nsum^{n-1}_{k=0}binom{n-1}{k}$$ and (2) from $$nsum^{n-1}_{k=0}binom{n-1}{k}$$ to $$ncdot2^{n-1}$$ respectively.






    sequences-and-series discrete-mathematics summation factorial







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      So, this is one of the questions in my textbook, which seems to be quite common: $$sum^{n}_{k=0}kcdot binom{n}{k}=ncdot2^{n-1}$$



      The same book provides the following solution:
      $$sum_{k=0}^{n}kcdot binom{n}{k}= sum^{n}_{k=0}ncdotbinom{n-1}{k-1}=nsum^{n-1}_{k=0}binom{n-1}{k}=ncdot2^{n-1}$$



      What is unclear to me, is (1) how I get from $$sum^{n}_{k=0}ncdotbinom{n-1}{k-1} $$ to$$nsum^{n-1}_{k=0}binom{n-1}{k}$$ and (2) from $$nsum^{n-1}_{k=0}binom{n-1}{k}$$ to $$ncdot2^{n-1}$$ respectively.






      sequences-and-series discrete-mathematics summation factorial







      share|cite|improve this question









      $endgroup$




      So, this is one of the questions in my textbook, which seems to be quite common: $$sum^{n}_{k=0}kcdot binom{n}{k}=ncdot2^{n-1}$$



      The same book provides the following solution:
      $$sum_{k=0}^{n}kcdot binom{n}{k}= sum^{n}_{k=0}ncdotbinom{n-1}{k-1}=nsum^{n-1}_{k=0}binom{n-1}{k}=ncdot2^{n-1}$$



      What is unclear to me, is (1) how I get from $$sum^{n}_{k=0}ncdotbinom{n-1}{k-1} $$ to$$nsum^{n-1}_{k=0}binom{n-1}{k}$$ and (2) from $$nsum^{n-1}_{k=0}binom{n-1}{k}$$ to $$ncdot2^{n-1}$$ respectively.






      sequences-and-series discrete-mathematics summation factorial




      sequences-and-series discrete-mathematics summation factorial






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      asked Dec 22 '18 at 15:26









      LuckyLucky

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      1127






















          2 Answers
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          $begingroup$

          (1) We have $$sum_{k=0}^nncdot{n-1choose k-1}=nsum_{k=0}^{n-1}{n-1choose k}$$ by a substitution $k'=k-1$. We can obviously place the $n$ at the front, and we then use ${n-1choose-1}=0$, so $k=0$ can be ignored.



          (2) Here we just use the formula $sum_{k=0}^n{nchoose k}=2^n$. There is a very nice combinatorial proof of this. The term ${nchoose k}$ represents the amount of ways to pick $k$ items out of $n$. If we sum over all $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items. There are $2^n$ ways, since we can choose for each item whether we pick it or not.



          There is also a very nice combinatorial proof of the statement you want to prove in the first place. The term $kcdot{nchoose k}$ can be thought of the amount of ways to pick $k$ items out of $n$ and then choosing one of the $k$ items as your favorite. Summing over $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items, and then choosing one of the picked items as your favorite. We can also reverse this: First choose your favorite item, and then pick any set of remaining items out of the $n-1$ left over items. This results in $ncdot2^{n-1}$ possibilities.






          share|cite|improve this answer









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            0












            $begingroup$

            The second part is very simple. As the $sum^{n-1}_{k=0}binom{n-1}{k}$ is the size of power set of a set with size of $n-1$. Hence, it is equal to $2^{n-1}$.



            Hint: For the first part, there is a mistake in the range of $sum$. If it would be true, you can show it using a simple change variable ($u = k - 1$ and change the range of sigma from ($1$ to $n$) to ($0$ to $n-1$)).






            share|cite|improve this answer











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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

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              active

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              active

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              1












              $begingroup$

              (1) We have $$sum_{k=0}^nncdot{n-1choose k-1}=nsum_{k=0}^{n-1}{n-1choose k}$$ by a substitution $k'=k-1$. We can obviously place the $n$ at the front, and we then use ${n-1choose-1}=0$, so $k=0$ can be ignored.



              (2) Here we just use the formula $sum_{k=0}^n{nchoose k}=2^n$. There is a very nice combinatorial proof of this. The term ${nchoose k}$ represents the amount of ways to pick $k$ items out of $n$. If we sum over all $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items. There are $2^n$ ways, since we can choose for each item whether we pick it or not.



              There is also a very nice combinatorial proof of the statement you want to prove in the first place. The term $kcdot{nchoose k}$ can be thought of the amount of ways to pick $k$ items out of $n$ and then choosing one of the $k$ items as your favorite. Summing over $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items, and then choosing one of the picked items as your favorite. We can also reverse this: First choose your favorite item, and then pick any set of remaining items out of the $n-1$ left over items. This results in $ncdot2^{n-1}$ possibilities.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                (1) We have $$sum_{k=0}^nncdot{n-1choose k-1}=nsum_{k=0}^{n-1}{n-1choose k}$$ by a substitution $k'=k-1$. We can obviously place the $n$ at the front, and we then use ${n-1choose-1}=0$, so $k=0$ can be ignored.



                (2) Here we just use the formula $sum_{k=0}^n{nchoose k}=2^n$. There is a very nice combinatorial proof of this. The term ${nchoose k}$ represents the amount of ways to pick $k$ items out of $n$. If we sum over all $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items. There are $2^n$ ways, since we can choose for each item whether we pick it or not.



                There is also a very nice combinatorial proof of the statement you want to prove in the first place. The term $kcdot{nchoose k}$ can be thought of the amount of ways to pick $k$ items out of $n$ and then choosing one of the $k$ items as your favorite. Summing over $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items, and then choosing one of the picked items as your favorite. We can also reverse this: First choose your favorite item, and then pick any set of remaining items out of the $n-1$ left over items. This results in $ncdot2^{n-1}$ possibilities.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  (1) We have $$sum_{k=0}^nncdot{n-1choose k-1}=nsum_{k=0}^{n-1}{n-1choose k}$$ by a substitution $k'=k-1$. We can obviously place the $n$ at the front, and we then use ${n-1choose-1}=0$, so $k=0$ can be ignored.



                  (2) Here we just use the formula $sum_{k=0}^n{nchoose k}=2^n$. There is a very nice combinatorial proof of this. The term ${nchoose k}$ represents the amount of ways to pick $k$ items out of $n$. If we sum over all $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items. There are $2^n$ ways, since we can choose for each item whether we pick it or not.



                  There is also a very nice combinatorial proof of the statement you want to prove in the first place. The term $kcdot{nchoose k}$ can be thought of the amount of ways to pick $k$ items out of $n$ and then choosing one of the $k$ items as your favorite. Summing over $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items, and then choosing one of the picked items as your favorite. We can also reverse this: First choose your favorite item, and then pick any set of remaining items out of the $n-1$ left over items. This results in $ncdot2^{n-1}$ possibilities.






                  share|cite|improve this answer









                  $endgroup$



                  (1) We have $$sum_{k=0}^nncdot{n-1choose k-1}=nsum_{k=0}^{n-1}{n-1choose k}$$ by a substitution $k'=k-1$. We can obviously place the $n$ at the front, and we then use ${n-1choose-1}=0$, so $k=0$ can be ignored.



                  (2) Here we just use the formula $sum_{k=0}^n{nchoose k}=2^n$. There is a very nice combinatorial proof of this. The term ${nchoose k}$ represents the amount of ways to pick $k$ items out of $n$. If we sum over all $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items. There are $2^n$ ways, since we can choose for each item whether we pick it or not.



                  There is also a very nice combinatorial proof of the statement you want to prove in the first place. The term $kcdot{nchoose k}$ can be thought of the amount of ways to pick $k$ items out of $n$ and then choosing one of the $k$ items as your favorite. Summing over $0leq kleq n$, we just look at the total amount of ways to pick any set out of $n$ items, and then choosing one of the picked items as your favorite. We can also reverse this: First choose your favorite item, and then pick any set of remaining items out of the $n-1$ left over items. This results in $ncdot2^{n-1}$ possibilities.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 22 '18 at 15:38









                  SmileyCraftSmileyCraft

                  3,566517




                  3,566517























                      0












                      $begingroup$

                      The second part is very simple. As the $sum^{n-1}_{k=0}binom{n-1}{k}$ is the size of power set of a set with size of $n-1$. Hence, it is equal to $2^{n-1}$.



                      Hint: For the first part, there is a mistake in the range of $sum$. If it would be true, you can show it using a simple change variable ($u = k - 1$ and change the range of sigma from ($1$ to $n$) to ($0$ to $n-1$)).






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        The second part is very simple. As the $sum^{n-1}_{k=0}binom{n-1}{k}$ is the size of power set of a set with size of $n-1$. Hence, it is equal to $2^{n-1}$.



                        Hint: For the first part, there is a mistake in the range of $sum$. If it would be true, you can show it using a simple change variable ($u = k - 1$ and change the range of sigma from ($1$ to $n$) to ($0$ to $n-1$)).






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          The second part is very simple. As the $sum^{n-1}_{k=0}binom{n-1}{k}$ is the size of power set of a set with size of $n-1$. Hence, it is equal to $2^{n-1}$.



                          Hint: For the first part, there is a mistake in the range of $sum$. If it would be true, you can show it using a simple change variable ($u = k - 1$ and change the range of sigma from ($1$ to $n$) to ($0$ to $n-1$)).






                          share|cite|improve this answer











                          $endgroup$



                          The second part is very simple. As the $sum^{n-1}_{k=0}binom{n-1}{k}$ is the size of power set of a set with size of $n-1$. Hence, it is equal to $2^{n-1}$.



                          Hint: For the first part, there is a mistake in the range of $sum$. If it would be true, you can show it using a simple change variable ($u = k - 1$ and change the range of sigma from ($1$ to $n$) to ($0$ to $n-1$)).







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 22 '18 at 15:34

























                          answered Dec 22 '18 at 15:29









                          OmGOmG

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                          2,512722






























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