Proving Sums or Products by Induction

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I am currently teaching Extension 1 Mathematics and seem to be coming unstuck on a few induction questions.
$$sum_k=1^nfrac1(2k-1)(2k+1)=fracn2n+1$$



First step is to prove the equation works for "1" - which it does. The second step is to prove the function works for $k+1$ with my working set out as follows:
$$LHS=sum_k=1^nfrac1(2k-1)(2k+1)+frac1(2n+1)(2n+3)$$
$$=fracn2n+1+frac1(2n+1)(2n+3)$$
$$=fracn(2n+3)+1(2n+1)(2n+3)$$



After setting out these steps - my text tells me the answer i'm looking for is $fracn+12n+3$ and I cant seem to find how to get from the final step of my working to the final answer! What am I missing here? (this is the second question I have come unstuck on a step before the final answer)






sequences-and-series induction education







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    up vote
    1
    down vote

    favorite












    I am currently teaching Extension 1 Mathematics and seem to be coming unstuck on a few induction questions.
    $$sum_k=1^nfrac1(2k-1)(2k+1)=fracn2n+1$$



    First step is to prove the equation works for "1" - which it does. The second step is to prove the function works for $k+1$ with my working set out as follows:
    $$LHS=sum_k=1^nfrac1(2k-1)(2k+1)+frac1(2n+1)(2n+3)$$
    $$=fracn2n+1+frac1(2n+1)(2n+3)$$
    $$=fracn(2n+3)+1(2n+1)(2n+3)$$



    After setting out these steps - my text tells me the answer i'm looking for is $fracn+12n+3$ and I cant seem to find how to get from the final step of my working to the final answer! What am I missing here? (this is the second question I have come unstuck on a step before the final answer)






    sequences-and-series induction education







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am currently teaching Extension 1 Mathematics and seem to be coming unstuck on a few induction questions.
      $$sum_k=1^nfrac1(2k-1)(2k+1)=fracn2n+1$$



      First step is to prove the equation works for "1" - which it does. The second step is to prove the function works for $k+1$ with my working set out as follows:
      $$LHS=sum_k=1^nfrac1(2k-1)(2k+1)+frac1(2n+1)(2n+3)$$
      $$=fracn2n+1+frac1(2n+1)(2n+3)$$
      $$=fracn(2n+3)+1(2n+1)(2n+3)$$



      After setting out these steps - my text tells me the answer i'm looking for is $fracn+12n+3$ and I cant seem to find how to get from the final step of my working to the final answer! What am I missing here? (this is the second question I have come unstuck on a step before the final answer)






      sequences-and-series induction education







      share|cite|improve this question















      I am currently teaching Extension 1 Mathematics and seem to be coming unstuck on a few induction questions.
      $$sum_k=1^nfrac1(2k-1)(2k+1)=fracn2n+1$$



      First step is to prove the equation works for "1" - which it does. The second step is to prove the function works for $k+1$ with my working set out as follows:
      $$LHS=sum_k=1^nfrac1(2k-1)(2k+1)+frac1(2n+1)(2n+3)$$
      $$=fracn2n+1+frac1(2n+1)(2n+3)$$
      $$=fracn(2n+3)+1(2n+1)(2n+3)$$



      After setting out these steps - my text tells me the answer i'm looking for is $fracn+12n+3$ and I cant seem to find how to get from the final step of my working to the final answer! What am I missing here? (this is the second question I have come unstuck on a step before the final answer)






      sequences-and-series induction education




      sequences-and-series induction education






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      share|cite|improve this question













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      edited 3 hours ago









      F.Carette

      1,07810




      1,07810










      asked 3 hours ago









      Holly Millican

      428




      428




















          3 Answers
          3






          active

          oldest

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          up vote
          2
          down vote



          accepted










          You're stuck at



          $$fracn(2n+3)+1colorred(2n+1)colorgreen(2n+3)$$



          And want to go to:



          $$fracn+1colorgreen2n+3$$



          So, you just have to expend and refactor the numerator and hope that a $(2n+1)$ factor will appear!



          $$n(2n+3)+1 = 2n^2+3n+1 = (n+1)(2n+1)$$



          Therefore:



          $$frack(2n+3)+1(2n+1)(2n+3)=frac(n+1)(2n+1)(2n+1)(2n+3)=fracn+12n+3$$






          share|cite|improve this answer






















          • Thank you, that was exactly what I was missing :)
            – Holly Millican
            3 hours ago

















          up vote
          2
          down vote













          We have $$frac2k^2+3k+1(2k+1)(2k+3)=frac(2k+1)(k+1)(2k+1)(2k+3) $$



          Note that the line on the right of LHS makes no sense, do not omit summation sign. Also be careful about the choice of notation, $n$ or $k$.






          share|cite|improve this answer




















          • Thankyou, I was missing the factorising
            – Holly Millican
            3 hours ago










          • I suggested an edit in the question to take care of your concerns ($n$/$k$ confusion and missing summation sign). In case it is accepted, you may want to modify your answer
            – F.Carette
            3 hours ago

















          up vote
          0
          down vote













          Fro the induction step we need to prove that



          $$sum_k=1^n+1frac1(2k-1)(2k+1)=frac1(2n+1)(2n+3)+sum_k=1^nfrac1(2k-1)(2k+1)stackrelInd. Hyp.=frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



          that is



          $$frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



          $$frac1(2n+1)(2n+3)=fracn+12n+3-fracn2n+1$$



          $$frac1(2n+1)(2n+3)=frac(2n+1)(n+1)-n(2n+3)(2n+1)(2n+3)$$



          $$frac1(2n+1)(2n+3)=fraccolorred2n^2+3n+1colorred-2n^2-3n(2n+1)(2n+3)=frac1(2n+1)(2n+3)$$






          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
            2
            down vote



            accepted










            You're stuck at



            $$fracn(2n+3)+1colorred(2n+1)colorgreen(2n+3)$$



            And want to go to:



            $$fracn+1colorgreen2n+3$$



            So, you just have to expend and refactor the numerator and hope that a $(2n+1)$ factor will appear!



            $$n(2n+3)+1 = 2n^2+3n+1 = (n+1)(2n+1)$$



            Therefore:



            $$frack(2n+3)+1(2n+1)(2n+3)=frac(n+1)(2n+1)(2n+1)(2n+3)=fracn+12n+3$$






            share|cite|improve this answer






















            • Thank you, that was exactly what I was missing :)
              – Holly Millican
              3 hours ago














            up vote
            2
            down vote



            accepted










            You're stuck at



            $$fracn(2n+3)+1colorred(2n+1)colorgreen(2n+3)$$



            And want to go to:



            $$fracn+1colorgreen2n+3$$



            So, you just have to expend and refactor the numerator and hope that a $(2n+1)$ factor will appear!



            $$n(2n+3)+1 = 2n^2+3n+1 = (n+1)(2n+1)$$



            Therefore:



            $$frack(2n+3)+1(2n+1)(2n+3)=frac(n+1)(2n+1)(2n+1)(2n+3)=fracn+12n+3$$






            share|cite|improve this answer






















            • Thank you, that was exactly what I was missing :)
              – Holly Millican
              3 hours ago












            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            You're stuck at



            $$fracn(2n+3)+1colorred(2n+1)colorgreen(2n+3)$$



            And want to go to:



            $$fracn+1colorgreen2n+3$$



            So, you just have to expend and refactor the numerator and hope that a $(2n+1)$ factor will appear!



            $$n(2n+3)+1 = 2n^2+3n+1 = (n+1)(2n+1)$$



            Therefore:



            $$frack(2n+3)+1(2n+1)(2n+3)=frac(n+1)(2n+1)(2n+1)(2n+3)=fracn+12n+3$$






            share|cite|improve this answer














            You're stuck at



            $$fracn(2n+3)+1colorred(2n+1)colorgreen(2n+3)$$



            And want to go to:



            $$fracn+1colorgreen2n+3$$



            So, you just have to expend and refactor the numerator and hope that a $(2n+1)$ factor will appear!



            $$n(2n+3)+1 = 2n^2+3n+1 = (n+1)(2n+1)$$



            Therefore:



            $$frack(2n+3)+1(2n+1)(2n+3)=frac(n+1)(2n+1)(2n+1)(2n+3)=fracn+12n+3$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 3 hours ago









            F.Carette

            1,07810




            1,07810











            • Thank you, that was exactly what I was missing :)
              – Holly Millican
              3 hours ago
















            • Thank you, that was exactly what I was missing :)
              – Holly Millican
              3 hours ago















            Thank you, that was exactly what I was missing :)
            – Holly Millican
            3 hours ago




            Thank you, that was exactly what I was missing :)
            – Holly Millican
            3 hours ago










            up vote
            2
            down vote













            We have $$frac2k^2+3k+1(2k+1)(2k+3)=frac(2k+1)(k+1)(2k+1)(2k+3) $$



            Note that the line on the right of LHS makes no sense, do not omit summation sign. Also be careful about the choice of notation, $n$ or $k$.






            share|cite|improve this answer




















            • Thankyou, I was missing the factorising
              – Holly Millican
              3 hours ago










            • I suggested an edit in the question to take care of your concerns ($n$/$k$ confusion and missing summation sign). In case it is accepted, you may want to modify your answer
              – F.Carette
              3 hours ago














            up vote
            2
            down vote













            We have $$frac2k^2+3k+1(2k+1)(2k+3)=frac(2k+1)(k+1)(2k+1)(2k+3) $$



            Note that the line on the right of LHS makes no sense, do not omit summation sign. Also be careful about the choice of notation, $n$ or $k$.






            share|cite|improve this answer




















            • Thankyou, I was missing the factorising
              – Holly Millican
              3 hours ago










            • I suggested an edit in the question to take care of your concerns ($n$/$k$ confusion and missing summation sign). In case it is accepted, you may want to modify your answer
              – F.Carette
              3 hours ago












            up vote
            2
            down vote










            up vote
            2
            down vote









            We have $$frac2k^2+3k+1(2k+1)(2k+3)=frac(2k+1)(k+1)(2k+1)(2k+3) $$



            Note that the line on the right of LHS makes no sense, do not omit summation sign. Also be careful about the choice of notation, $n$ or $k$.






            share|cite|improve this answer












            We have $$frac2k^2+3k+1(2k+1)(2k+3)=frac(2k+1)(k+1)(2k+1)(2k+3) $$



            Note that the line on the right of LHS makes no sense, do not omit summation sign. Also be careful about the choice of notation, $n$ or $k$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            Siong Thye Goh

            88.2k1460111




            88.2k1460111











            • Thankyou, I was missing the factorising
              – Holly Millican
              3 hours ago










            • I suggested an edit in the question to take care of your concerns ($n$/$k$ confusion and missing summation sign). In case it is accepted, you may want to modify your answer
              – F.Carette
              3 hours ago
















            • Thankyou, I was missing the factorising
              – Holly Millican
              3 hours ago










            • I suggested an edit in the question to take care of your concerns ($n$/$k$ confusion and missing summation sign). In case it is accepted, you may want to modify your answer
              – F.Carette
              3 hours ago















            Thankyou, I was missing the factorising
            – Holly Millican
            3 hours ago




            Thankyou, I was missing the factorising
            – Holly Millican
            3 hours ago












            I suggested an edit in the question to take care of your concerns ($n$/$k$ confusion and missing summation sign). In case it is accepted, you may want to modify your answer
            – F.Carette
            3 hours ago




            I suggested an edit in the question to take care of your concerns ($n$/$k$ confusion and missing summation sign). In case it is accepted, you may want to modify your answer
            – F.Carette
            3 hours ago










            up vote
            0
            down vote













            Fro the induction step we need to prove that



            $$sum_k=1^n+1frac1(2k-1)(2k+1)=frac1(2n+1)(2n+3)+sum_k=1^nfrac1(2k-1)(2k+1)stackrelInd. Hyp.=frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



            that is



            $$frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



            $$frac1(2n+1)(2n+3)=fracn+12n+3-fracn2n+1$$



            $$frac1(2n+1)(2n+3)=frac(2n+1)(n+1)-n(2n+3)(2n+1)(2n+3)$$



            $$frac1(2n+1)(2n+3)=fraccolorred2n^2+3n+1colorred-2n^2-3n(2n+1)(2n+3)=frac1(2n+1)(2n+3)$$






            share|cite|improve this answer


























              up vote
              0
              down vote













              Fro the induction step we need to prove that



              $$sum_k=1^n+1frac1(2k-1)(2k+1)=frac1(2n+1)(2n+3)+sum_k=1^nfrac1(2k-1)(2k+1)stackrelInd. Hyp.=frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



              that is



              $$frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



              $$frac1(2n+1)(2n+3)=fracn+12n+3-fracn2n+1$$



              $$frac1(2n+1)(2n+3)=frac(2n+1)(n+1)-n(2n+3)(2n+1)(2n+3)$$



              $$frac1(2n+1)(2n+3)=fraccolorred2n^2+3n+1colorred-2n^2-3n(2n+1)(2n+3)=frac1(2n+1)(2n+3)$$






              share|cite|improve this answer
























                up vote
                0
                down vote










                up vote
                0
                down vote









                Fro the induction step we need to prove that



                $$sum_k=1^n+1frac1(2k-1)(2k+1)=frac1(2n+1)(2n+3)+sum_k=1^nfrac1(2k-1)(2k+1)stackrelInd. Hyp.=frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



                that is



                $$frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



                $$frac1(2n+1)(2n+3)=fracn+12n+3-fracn2n+1$$



                $$frac1(2n+1)(2n+3)=frac(2n+1)(n+1)-n(2n+3)(2n+1)(2n+3)$$



                $$frac1(2n+1)(2n+3)=fraccolorred2n^2+3n+1colorred-2n^2-3n(2n+1)(2n+3)=frac1(2n+1)(2n+3)$$






                share|cite|improve this answer














                Fro the induction step we need to prove that



                $$sum_k=1^n+1frac1(2k-1)(2k+1)=frac1(2n+1)(2n+3)+sum_k=1^nfrac1(2k-1)(2k+1)stackrelInd. Hyp.=frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



                that is



                $$frac1(2n+1)(2n+3)+fracn2n+1stackrel?=fracn+12n+3$$



                $$frac1(2n+1)(2n+3)=fracn+12n+3-fracn2n+1$$



                $$frac1(2n+1)(2n+3)=frac(2n+1)(n+1)-n(2n+3)(2n+1)(2n+3)$$



                $$frac1(2n+1)(2n+3)=fraccolorred2n^2+3n+1colorred-2n^2-3n(2n+1)(2n+3)=frac1(2n+1)(2n+3)$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 3 hours ago

























                answered 3 hours ago









                gimusi

                78.4k73889




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