Using the binomial expansion to show $sum_{k=0}^n {n choose k} frac{1}{2^k} = (3/2)^n$












1












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I have to show $$sum_{k=0}^n {n choose k} frac{1}{2^k} = (3/2)^n$$ using the binomial theorem. I haven't had any practice on these types of questions so i'm unsure as to how to proceed. Would anybody be able to what the method is for solving these types of questions, and, is there any general method of what we're trying to do? What's the thought process to solve these types of questions



Much appreciated.
Thank you






binomial-theorem







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  • 1




    $begingroup$
    Use induction on $n$.
    $endgroup$
    – Wuestenfux
    Jan 2 at 13:24










  • $begingroup$
    Hint: $3/2=1+1/2$
    $endgroup$
    – Wojowu
    Jan 2 at 13:25










  • $begingroup$
    Proof by induction isn't on my syllabus. How do i do it without it?
    $endgroup$
    – Young's Modulus
    Jan 2 at 13:25










  • $begingroup$
    Hint ( a little more than @Wojowu ): Do you know how to expand $(1 + x)^n$?
    $endgroup$
    – Ethan Bolker
    Jan 2 at 13:26
















1












$begingroup$


I have to show $$sum_{k=0}^n {n choose k} frac{1}{2^k} = (3/2)^n$$ using the binomial theorem. I haven't had any practice on these types of questions so i'm unsure as to how to proceed. Would anybody be able to what the method is for solving these types of questions, and, is there any general method of what we're trying to do? What's the thought process to solve these types of questions



Much appreciated.
Thank you






binomial-theorem







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Use induction on $n$.
    $endgroup$
    – Wuestenfux
    Jan 2 at 13:24










  • $begingroup$
    Hint: $3/2=1+1/2$
    $endgroup$
    – Wojowu
    Jan 2 at 13:25










  • $begingroup$
    Proof by induction isn't on my syllabus. How do i do it without it?
    $endgroup$
    – Young's Modulus
    Jan 2 at 13:25










  • $begingroup$
    Hint ( a little more than @Wojowu ): Do you know how to expand $(1 + x)^n$?
    $endgroup$
    – Ethan Bolker
    Jan 2 at 13:26














1












1








1





$begingroup$


I have to show $$sum_{k=0}^n {n choose k} frac{1}{2^k} = (3/2)^n$$ using the binomial theorem. I haven't had any practice on these types of questions so i'm unsure as to how to proceed. Would anybody be able to what the method is for solving these types of questions, and, is there any general method of what we're trying to do? What's the thought process to solve these types of questions



Much appreciated.
Thank you






binomial-theorem







share|cite|improve this question









$endgroup$




I have to show $$sum_{k=0}^n {n choose k} frac{1}{2^k} = (3/2)^n$$ using the binomial theorem. I haven't had any practice on these types of questions so i'm unsure as to how to proceed. Would anybody be able to what the method is for solving these types of questions, and, is there any general method of what we're trying to do? What's the thought process to solve these types of questions



Much appreciated.
Thank you






binomial-theorem




binomial-theorem






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share|cite|improve this question











share|cite|improve this question




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asked Jan 2 at 13:23









Young's ModulusYoung's Modulus

61




61








  • 1




    $begingroup$
    Use induction on $n$.
    $endgroup$
    – Wuestenfux
    Jan 2 at 13:24










  • $begingroup$
    Hint: $3/2=1+1/2$
    $endgroup$
    – Wojowu
    Jan 2 at 13:25










  • $begingroup$
    Proof by induction isn't on my syllabus. How do i do it without it?
    $endgroup$
    – Young's Modulus
    Jan 2 at 13:25










  • $begingroup$
    Hint ( a little more than @Wojowu ): Do you know how to expand $(1 + x)^n$?
    $endgroup$
    – Ethan Bolker
    Jan 2 at 13:26














  • 1




    $begingroup$
    Use induction on $n$.
    $endgroup$
    – Wuestenfux
    Jan 2 at 13:24










  • $begingroup$
    Hint: $3/2=1+1/2$
    $endgroup$
    – Wojowu
    Jan 2 at 13:25










  • $begingroup$
    Proof by induction isn't on my syllabus. How do i do it without it?
    $endgroup$
    – Young's Modulus
    Jan 2 at 13:25










  • $begingroup$
    Hint ( a little more than @Wojowu ): Do you know how to expand $(1 + x)^n$?
    $endgroup$
    – Ethan Bolker
    Jan 2 at 13:26








1




1




$begingroup$
Use induction on $n$.
$endgroup$
– Wuestenfux
Jan 2 at 13:24




$begingroup$
Use induction on $n$.
$endgroup$
– Wuestenfux
Jan 2 at 13:24












$begingroup$
Hint: $3/2=1+1/2$
$endgroup$
– Wojowu
Jan 2 at 13:25




$begingroup$
Hint: $3/2=1+1/2$
$endgroup$
– Wojowu
Jan 2 at 13:25












$begingroup$
Proof by induction isn't on my syllabus. How do i do it without it?
$endgroup$
– Young's Modulus
Jan 2 at 13:25




$begingroup$
Proof by induction isn't on my syllabus. How do i do it without it?
$endgroup$
– Young's Modulus
Jan 2 at 13:25












$begingroup$
Hint ( a little more than @Wojowu ): Do you know how to expand $(1 + x)^n$?
$endgroup$
– Ethan Bolker
Jan 2 at 13:26




$begingroup$
Hint ( a little more than @Wojowu ): Do you know how to expand $(1 + x)^n$?
$endgroup$
– Ethan Bolker
Jan 2 at 13:26










4 Answers
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$begingroup$

$left(dfrac{3}{2}right)^n= left(dfrac{1}{2}+1right)^n =sumlimits_{k=0}^ndbinom{n}{k}dfrac{1}{2^k}$






share|cite|improve this answer









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    2












    $begingroup$

    Take $x=1/2$ and $y=1$ in the expansion
    $$
    (x+y)^n= sum_{k=0}^n binom{n}{k}x^k y^{n-k}.
    $$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      $$(3/2)^n=(1+1/2)^n=sum_{k=0}^n {n choose k} frac{1}{2^k}.$$
      Last equation follows from binomial theorem.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Right, Of course, I somehow managed to overthink this. Thanks.
        $endgroup$
        – Young's Modulus
        Jan 2 at 13:35



















      1












      $begingroup$

      There is a theorem (Binomial theorem, demostrable by induction) that shows you:
      begin{equation}
      (a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}
      end{equation}
      Then if you use the $displaystyle a=frac{1}{2}$ and $b=1$, you obtain:
      $$sum_{k=0}^n binom{n}{k} frac{1}{2^k} = (3/2)^n$$
      Proof:
      The demostration of the Binomial Theorem is not complicated, the case $n=1$ is obviously true, then if we assume the case $n$, and we came to prove the case $n+1$ and the theorem is proven. If:



      $$(a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}$$
      Then:
      begin{eqnarray*}
      (a+b)^{n+1} &=& (a+b)(a+b)^{n} = a(a+b)^{n} + b(a+b)^{n} \
      (a+b)^{n+1} &=& asum_{k=0}^n binom{n}{k} a^k b^{n-k} + bsum_{k=0}^n binom{n}{k} a^k b^{n-k} \
      (a+b)^{n+1} &=& sum_{k=0}^n binom{n}{k} a^{k+1} b^{n-k} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
      (a+b)^{n+1} &=& sum_{k=1}^{n+1} binom{n}{k-1} a^{k} b^{n-k+1} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
      (a+b)^{n+1} &=& left[a^{n+1} + sum_{k=1}^{n} binom{n}{k-1} a^{k} b^{n-k+1} right] + left[b^{n+1} +sum_{k=1}^n binom{n}{k} a^k b^{n-k+1} right] \
      (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ binom{n}{k-1} + binom{n}{k}right]right) \
      (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!}{(k-1)!(n-k+1)!} + frac{n!}{k!(n-k)!}right]right) \
      (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!k}{k!(n-k+1)!} + frac{n!(n-k+1)}{k!(n-k+1)!}right]right) \
      (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!(n+1)}{k!(n-k+1)!}right]right) \
      (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left[a^k b^{n-k+1} frac{(n+1)!}{k!(n+1-k)!}right] \
      (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}binom{n+1}{k}a^k b^{n+1-k} \
      (a+b)^{n+1} &=& sum_{k=0}^{n+1}binom{n+1}{k}a^k b^{n+1-k} \
      end{eqnarray*}



      And now the theorem is demostrated by induction. Here is a video of the demonstration.






      share|cite|improve this answer











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        4 Answers
        4






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        4 Answers
        4






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

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        3












        $begingroup$

        $left(dfrac{3}{2}right)^n= left(dfrac{1}{2}+1right)^n =sumlimits_{k=0}^ndbinom{n}{k}dfrac{1}{2^k}$






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          $left(dfrac{3}{2}right)^n= left(dfrac{1}{2}+1right)^n =sumlimits_{k=0}^ndbinom{n}{k}dfrac{1}{2^k}$






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            $left(dfrac{3}{2}right)^n= left(dfrac{1}{2}+1right)^n =sumlimits_{k=0}^ndbinom{n}{k}dfrac{1}{2^k}$






            share|cite|improve this answer









            $endgroup$



            $left(dfrac{3}{2}right)^n= left(dfrac{1}{2}+1right)^n =sumlimits_{k=0}^ndbinom{n}{k}dfrac{1}{2^k}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 2 at 13:30









            Martín Vacas VignoloMartín Vacas Vignolo

            3,811623




            3,811623























                2












                $begingroup$

                Take $x=1/2$ and $y=1$ in the expansion
                $$
                (x+y)^n= sum_{k=0}^n binom{n}{k}x^k y^{n-k}.
                $$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Take $x=1/2$ and $y=1$ in the expansion
                  $$
                  (x+y)^n= sum_{k=0}^n binom{n}{k}x^k y^{n-k}.
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Take $x=1/2$ and $y=1$ in the expansion
                    $$
                    (x+y)^n= sum_{k=0}^n binom{n}{k}x^k y^{n-k}.
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    Take $x=1/2$ and $y=1$ in the expansion
                    $$
                    (x+y)^n= sum_{k=0}^n binom{n}{k}x^k y^{n-k}.
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 2 at 13:27









                    BerkheimerBerkheimer

                    1,437924




                    1,437924























                        2












                        $begingroup$

                        $$(3/2)^n=(1+1/2)^n=sum_{k=0}^n {n choose k} frac{1}{2^k}.$$
                        Last equation follows from binomial theorem.






                        share|cite|improve this answer









                        $endgroup$









                        • 1




                          $begingroup$
                          Right, Of course, I somehow managed to overthink this. Thanks.
                          $endgroup$
                          – Young's Modulus
                          Jan 2 at 13:35
















                        2












                        $begingroup$

                        $$(3/2)^n=(1+1/2)^n=sum_{k=0}^n {n choose k} frac{1}{2^k}.$$
                        Last equation follows from binomial theorem.






                        share|cite|improve this answer









                        $endgroup$









                        • 1




                          $begingroup$
                          Right, Of course, I somehow managed to overthink this. Thanks.
                          $endgroup$
                          – Young's Modulus
                          Jan 2 at 13:35














                        2












                        2








                        2





                        $begingroup$

                        $$(3/2)^n=(1+1/2)^n=sum_{k=0}^n {n choose k} frac{1}{2^k}.$$
                        Last equation follows from binomial theorem.






                        share|cite|improve this answer









                        $endgroup$



                        $$(3/2)^n=(1+1/2)^n=sum_{k=0}^n {n choose k} frac{1}{2^k}.$$
                        Last equation follows from binomial theorem.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 2 at 13:29









                        Thomas ShelbyThomas Shelby

                        2,119220




                        2,119220








                        • 1




                          $begingroup$
                          Right, Of course, I somehow managed to overthink this. Thanks.
                          $endgroup$
                          – Young's Modulus
                          Jan 2 at 13:35














                        • 1




                          $begingroup$
                          Right, Of course, I somehow managed to overthink this. Thanks.
                          $endgroup$
                          – Young's Modulus
                          Jan 2 at 13:35








                        1




                        1




                        $begingroup$
                        Right, Of course, I somehow managed to overthink this. Thanks.
                        $endgroup$
                        – Young's Modulus
                        Jan 2 at 13:35




                        $begingroup$
                        Right, Of course, I somehow managed to overthink this. Thanks.
                        $endgroup$
                        – Young's Modulus
                        Jan 2 at 13:35











                        1












                        $begingroup$

                        There is a theorem (Binomial theorem, demostrable by induction) that shows you:
                        begin{equation}
                        (a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}
                        end{equation}
                        Then if you use the $displaystyle a=frac{1}{2}$ and $b=1$, you obtain:
                        $$sum_{k=0}^n binom{n}{k} frac{1}{2^k} = (3/2)^n$$
                        Proof:
                        The demostration of the Binomial Theorem is not complicated, the case $n=1$ is obviously true, then if we assume the case $n$, and we came to prove the case $n+1$ and the theorem is proven. If:



                        $$(a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}$$
                        Then:
                        begin{eqnarray*}
                        (a+b)^{n+1} &=& (a+b)(a+b)^{n} = a(a+b)^{n} + b(a+b)^{n} \
                        (a+b)^{n+1} &=& asum_{k=0}^n binom{n}{k} a^k b^{n-k} + bsum_{k=0}^n binom{n}{k} a^k b^{n-k} \
                        (a+b)^{n+1} &=& sum_{k=0}^n binom{n}{k} a^{k+1} b^{n-k} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
                        (a+b)^{n+1} &=& sum_{k=1}^{n+1} binom{n}{k-1} a^{k} b^{n-k+1} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
                        (a+b)^{n+1} &=& left[a^{n+1} + sum_{k=1}^{n} binom{n}{k-1} a^{k} b^{n-k+1} right] + left[b^{n+1} +sum_{k=1}^n binom{n}{k} a^k b^{n-k+1} right] \
                        (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ binom{n}{k-1} + binom{n}{k}right]right) \
                        (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!}{(k-1)!(n-k+1)!} + frac{n!}{k!(n-k)!}right]right) \
                        (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!k}{k!(n-k+1)!} + frac{n!(n-k+1)}{k!(n-k+1)!}right]right) \
                        (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!(n+1)}{k!(n-k+1)!}right]right) \
                        (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left[a^k b^{n-k+1} frac{(n+1)!}{k!(n+1-k)!}right] \
                        (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}binom{n+1}{k}a^k b^{n+1-k} \
                        (a+b)^{n+1} &=& sum_{k=0}^{n+1}binom{n+1}{k}a^k b^{n+1-k} \
                        end{eqnarray*}



                        And now the theorem is demostrated by induction. Here is a video of the demonstration.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          There is a theorem (Binomial theorem, demostrable by induction) that shows you:
                          begin{equation}
                          (a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}
                          end{equation}
                          Then if you use the $displaystyle a=frac{1}{2}$ and $b=1$, you obtain:
                          $$sum_{k=0}^n binom{n}{k} frac{1}{2^k} = (3/2)^n$$
                          Proof:
                          The demostration of the Binomial Theorem is not complicated, the case $n=1$ is obviously true, then if we assume the case $n$, and we came to prove the case $n+1$ and the theorem is proven. If:



                          $$(a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}$$
                          Then:
                          begin{eqnarray*}
                          (a+b)^{n+1} &=& (a+b)(a+b)^{n} = a(a+b)^{n} + b(a+b)^{n} \
                          (a+b)^{n+1} &=& asum_{k=0}^n binom{n}{k} a^k b^{n-k} + bsum_{k=0}^n binom{n}{k} a^k b^{n-k} \
                          (a+b)^{n+1} &=& sum_{k=0}^n binom{n}{k} a^{k+1} b^{n-k} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
                          (a+b)^{n+1} &=& sum_{k=1}^{n+1} binom{n}{k-1} a^{k} b^{n-k+1} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
                          (a+b)^{n+1} &=& left[a^{n+1} + sum_{k=1}^{n} binom{n}{k-1} a^{k} b^{n-k+1} right] + left[b^{n+1} +sum_{k=1}^n binom{n}{k} a^k b^{n-k+1} right] \
                          (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ binom{n}{k-1} + binom{n}{k}right]right) \
                          (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!}{(k-1)!(n-k+1)!} + frac{n!}{k!(n-k)!}right]right) \
                          (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!k}{k!(n-k+1)!} + frac{n!(n-k+1)}{k!(n-k+1)!}right]right) \
                          (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!(n+1)}{k!(n-k+1)!}right]right) \
                          (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left[a^k b^{n-k+1} frac{(n+1)!}{k!(n+1-k)!}right] \
                          (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}binom{n+1}{k}a^k b^{n+1-k} \
                          (a+b)^{n+1} &=& sum_{k=0}^{n+1}binom{n+1}{k}a^k b^{n+1-k} \
                          end{eqnarray*}



                          And now the theorem is demostrated by induction. Here is a video of the demonstration.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            There is a theorem (Binomial theorem, demostrable by induction) that shows you:
                            begin{equation}
                            (a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}
                            end{equation}
                            Then if you use the $displaystyle a=frac{1}{2}$ and $b=1$, you obtain:
                            $$sum_{k=0}^n binom{n}{k} frac{1}{2^k} = (3/2)^n$$
                            Proof:
                            The demostration of the Binomial Theorem is not complicated, the case $n=1$ is obviously true, then if we assume the case $n$, and we came to prove the case $n+1$ and the theorem is proven. If:



                            $$(a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}$$
                            Then:
                            begin{eqnarray*}
                            (a+b)^{n+1} &=& (a+b)(a+b)^{n} = a(a+b)^{n} + b(a+b)^{n} \
                            (a+b)^{n+1} &=& asum_{k=0}^n binom{n}{k} a^k b^{n-k} + bsum_{k=0}^n binom{n}{k} a^k b^{n-k} \
                            (a+b)^{n+1} &=& sum_{k=0}^n binom{n}{k} a^{k+1} b^{n-k} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
                            (a+b)^{n+1} &=& sum_{k=1}^{n+1} binom{n}{k-1} a^{k} b^{n-k+1} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
                            (a+b)^{n+1} &=& left[a^{n+1} + sum_{k=1}^{n} binom{n}{k-1} a^{k} b^{n-k+1} right] + left[b^{n+1} +sum_{k=1}^n binom{n}{k} a^k b^{n-k+1} right] \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ binom{n}{k-1} + binom{n}{k}right]right) \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!}{(k-1)!(n-k+1)!} + frac{n!}{k!(n-k)!}right]right) \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!k}{k!(n-k+1)!} + frac{n!(n-k+1)}{k!(n-k+1)!}right]right) \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!(n+1)}{k!(n-k+1)!}right]right) \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left[a^k b^{n-k+1} frac{(n+1)!}{k!(n+1-k)!}right] \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}binom{n+1}{k}a^k b^{n+1-k} \
                            (a+b)^{n+1} &=& sum_{k=0}^{n+1}binom{n+1}{k}a^k b^{n+1-k} \
                            end{eqnarray*}



                            And now the theorem is demostrated by induction. Here is a video of the demonstration.






                            share|cite|improve this answer











                            $endgroup$



                            There is a theorem (Binomial theorem, demostrable by induction) that shows you:
                            begin{equation}
                            (a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}
                            end{equation}
                            Then if you use the $displaystyle a=frac{1}{2}$ and $b=1$, you obtain:
                            $$sum_{k=0}^n binom{n}{k} frac{1}{2^k} = (3/2)^n$$
                            Proof:
                            The demostration of the Binomial Theorem is not complicated, the case $n=1$ is obviously true, then if we assume the case $n$, and we came to prove the case $n+1$ and the theorem is proven. If:



                            $$(a+b)^n = sum_{k=0}^n binom{n}{k} a^k b^{n-k}$$
                            Then:
                            begin{eqnarray*}
                            (a+b)^{n+1} &=& (a+b)(a+b)^{n} = a(a+b)^{n} + b(a+b)^{n} \
                            (a+b)^{n+1} &=& asum_{k=0}^n binom{n}{k} a^k b^{n-k} + bsum_{k=0}^n binom{n}{k} a^k b^{n-k} \
                            (a+b)^{n+1} &=& sum_{k=0}^n binom{n}{k} a^{k+1} b^{n-k} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
                            (a+b)^{n+1} &=& sum_{k=1}^{n+1} binom{n}{k-1} a^{k} b^{n-k+1} + sum_{k=0}^n binom{n}{k} a^k b^{n-k+1} \
                            (a+b)^{n+1} &=& left[a^{n+1} + sum_{k=1}^{n} binom{n}{k-1} a^{k} b^{n-k+1} right] + left[b^{n+1} +sum_{k=1}^n binom{n}{k} a^k b^{n-k+1} right] \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ binom{n}{k-1} + binom{n}{k}right]right) \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!}{(k-1)!(n-k+1)!} + frac{n!}{k!(n-k)!}right]right) \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!k}{k!(n-k+1)!} + frac{n!(n-k+1)}{k!(n-k+1)!}right]right) \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left(a^k b^{n-k+1}left[ frac{n!(n+1)}{k!(n-k+1)!}right]right) \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}left[a^k b^{n-k+1} frac{(n+1)!}{k!(n+1-k)!}right] \
                            (a+b)^{n+1} &=& a^{n+1} + b^{n+1} + sum_{k=1}^{n}binom{n+1}{k}a^k b^{n+1-k} \
                            (a+b)^{n+1} &=& sum_{k=0}^{n+1}binom{n+1}{k}a^k b^{n+1-k} \
                            end{eqnarray*}



                            And now the theorem is demostrated by induction. Here is a video of the demonstration.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 10 at 0:18

























                            answered Jan 2 at 13:31









                            El PastaEl Pasta

                            34515




                            34515






























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