How do i do this mathematical induction question?











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My question:$5+10+20+...+5(2)^{n-1} = 5(2^n -1)$




  1. So first step i have to prove LHS = RHS when $n=1$, which is true.

  2. Then I assume the statement is true for $n=k$

  3. Since the statement is true for $n=k$ then for $n=k+1$


My workings:



$5+10+20+...+5(2)^{k-1} +5(2)^{(k+1)-1}= 5(2^{k+1} -1)$



LHS: $5(2^{k-1}) + 5(2)^k$



Then I do not know how to proceed to simplify, in general, can someone show some steps and show me how to tackle simplifying this kind of questions?










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  • LHS is $5(2^k-1)+5 cdot2^k$!
    – Jens Schwaiger
    yesterday












  • I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
    – maxmilgram
    yesterday

















up vote
0
down vote

favorite












My question:$5+10+20+...+5(2)^{n-1} = 5(2^n -1)$




  1. So first step i have to prove LHS = RHS when $n=1$, which is true.

  2. Then I assume the statement is true for $n=k$

  3. Since the statement is true for $n=k$ then for $n=k+1$


My workings:



$5+10+20+...+5(2)^{k-1} +5(2)^{(k+1)-1}= 5(2^{k+1} -1)$



LHS: $5(2^{k-1}) + 5(2)^k$



Then I do not know how to proceed to simplify, in general, can someone show some steps and show me how to tackle simplifying this kind of questions?










share|cite|improve this question






















  • LHS is $5(2^k-1)+5 cdot2^k$!
    – Jens Schwaiger
    yesterday












  • I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
    – maxmilgram
    yesterday















up vote
0
down vote

favorite









up vote
0
down vote

favorite











My question:$5+10+20+...+5(2)^{n-1} = 5(2^n -1)$




  1. So first step i have to prove LHS = RHS when $n=1$, which is true.

  2. Then I assume the statement is true for $n=k$

  3. Since the statement is true for $n=k$ then for $n=k+1$


My workings:



$5+10+20+...+5(2)^{k-1} +5(2)^{(k+1)-1}= 5(2^{k+1} -1)$



LHS: $5(2^{k-1}) + 5(2)^k$



Then I do not know how to proceed to simplify, in general, can someone show some steps and show me how to tackle simplifying this kind of questions?










share|cite|improve this question













My question:$5+10+20+...+5(2)^{n-1} = 5(2^n -1)$




  1. So first step i have to prove LHS = RHS when $n=1$, which is true.

  2. Then I assume the statement is true for $n=k$

  3. Since the statement is true for $n=k$ then for $n=k+1$


My workings:



$5+10+20+...+5(2)^{k-1} +5(2)^{(k+1)-1}= 5(2^{k+1} -1)$



LHS: $5(2^{k-1}) + 5(2)^k$



Then I do not know how to proceed to simplify, in general, can someone show some steps and show me how to tackle simplifying this kind of questions?







algebra-precalculus induction exponential-function






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asked yesterday









Tfue

486




486












  • LHS is $5(2^k-1)+5 cdot2^k$!
    – Jens Schwaiger
    yesterday












  • I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
    – maxmilgram
    yesterday




















  • LHS is $5(2^k-1)+5 cdot2^k$!
    – Jens Schwaiger
    yesterday












  • I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
    – maxmilgram
    yesterday


















LHS is $5(2^k-1)+5 cdot2^k$!
– Jens Schwaiger
yesterday






LHS is $5(2^k-1)+5 cdot2^k$!
– Jens Schwaiger
yesterday














I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
– maxmilgram
yesterday






I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
– maxmilgram
yesterday












3 Answers
3






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up vote
1
down vote



accepted










$5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$






share|cite|improve this answer





















  • how? what did you do?
    – Tfue
    yesterday










  • yo all g, i just need to get me eyes check!
    – Tfue
    yesterday


















up vote
0
down vote













For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
=5(2^k+1+2^{k+1})=5(2^{k+2}-1)$



So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.






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    Well, induction is not necessary.
    The equation can be simplified by canceling $5$ from both sides.
    We are left with
    $$1+2+4..2^{n-1}=2^n -1$$
    The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
    The sum $S_n$ is
    $$S_n= frac{a(r^n -1)}{r-1}$$
    $$=frac{(1)(2^n -1)}{2-1}$$
    $$=2^n -1$$
    Hence proved.
    Hope this helps.






    share|cite|improve this answer








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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      $5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$






      share|cite|improve this answer





















      • how? what did you do?
        – Tfue
        yesterday










      • yo all g, i just need to get me eyes check!
        – Tfue
        yesterday















      up vote
      1
      down vote



      accepted










      $5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$






      share|cite|improve this answer





















      • how? what did you do?
        – Tfue
        yesterday










      • yo all g, i just need to get me eyes check!
        – Tfue
        yesterday













      up vote
      1
      down vote



      accepted







      up vote
      1
      down vote



      accepted






      $5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$






      share|cite|improve this answer












      $5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered yesterday









      MoonKnight

      1,189510




      1,189510












      • how? what did you do?
        – Tfue
        yesterday










      • yo all g, i just need to get me eyes check!
        – Tfue
        yesterday


















      • how? what did you do?
        – Tfue
        yesterday










      • yo all g, i just need to get me eyes check!
        – Tfue
        yesterday
















      how? what did you do?
      – Tfue
      yesterday




      how? what did you do?
      – Tfue
      yesterday












      yo all g, i just need to get me eyes check!
      – Tfue
      yesterday




      yo all g, i just need to get me eyes check!
      – Tfue
      yesterday










      up vote
      0
      down vote













      For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
      =5(2^k+1+2^{k+1})=5(2^{k+2}-1)$



      So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.






      share|cite|improve this answer

























        up vote
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        down vote













        For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
        =5(2^k+1+2^{k+1})=5(2^{k+2}-1)$



        So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
          =5(2^k+1+2^{k+1})=5(2^{k+2}-1)$



          So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.






          share|cite|improve this answer












          For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
          =5(2^k+1+2^{k+1})=5(2^{k+2}-1)$



          So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Yadati Kiran

          231110




          231110






















              up vote
              0
              down vote













              Well, induction is not necessary.
              The equation can be simplified by canceling $5$ from both sides.
              We are left with
              $$1+2+4..2^{n-1}=2^n -1$$
              The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
              The sum $S_n$ is
              $$S_n= frac{a(r^n -1)}{r-1}$$
              $$=frac{(1)(2^n -1)}{2-1}$$
              $$=2^n -1$$
              Hence proved.
              Hope this helps.






              share|cite|improve this answer








              New contributor




              AryanSonwatikar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






















                up vote
                0
                down vote













                Well, induction is not necessary.
                The equation can be simplified by canceling $5$ from both sides.
                We are left with
                $$1+2+4..2^{n-1}=2^n -1$$
                The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
                The sum $S_n$ is
                $$S_n= frac{a(r^n -1)}{r-1}$$
                $$=frac{(1)(2^n -1)}{2-1}$$
                $$=2^n -1$$
                Hence proved.
                Hope this helps.






                share|cite|improve this answer








                New contributor




                AryanSonwatikar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.




















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Well, induction is not necessary.
                  The equation can be simplified by canceling $5$ from both sides.
                  We are left with
                  $$1+2+4..2^{n-1}=2^n -1$$
                  The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
                  The sum $S_n$ is
                  $$S_n= frac{a(r^n -1)}{r-1}$$
                  $$=frac{(1)(2^n -1)}{2-1}$$
                  $$=2^n -1$$
                  Hence proved.
                  Hope this helps.






                  share|cite|improve this answer








                  New contributor




                  AryanSonwatikar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  Well, induction is not necessary.
                  The equation can be simplified by canceling $5$ from both sides.
                  We are left with
                  $$1+2+4..2^{n-1}=2^n -1$$
                  The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
                  The sum $S_n$ is
                  $$S_n= frac{a(r^n -1)}{r-1}$$
                  $$=frac{(1)(2^n -1)}{2-1}$$
                  $$=2^n -1$$
                  Hence proved.
                  Hope this helps.







                  share|cite|improve this answer








                  New contributor




                  AryanSonwatikar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  AryanSonwatikar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered 45 mins ago









                  AryanSonwatikar

                  195




                  195




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                  New contributor





                  AryanSonwatikar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  AryanSonwatikar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






























                       

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