How to find the $n-th$ term of the series $3,7,12,18…$












1












$begingroup$


My attempt:




Let,$$space space space space space space space S=3+7+12+18+......+t_n$$
$$space space space space space space space spacespace space space space space space space space space S=space space space space space space space space 3+7+12+18+...t_{n-1}+t_n$$




Subtracting,
$$0=3+4+5+6+...(t_n-t_{n-1})-t_n$$
$$implies t_n=frac{n(n+1)}{2}-3$$
By adding and subtracting $1,2$.
However this gives a negative value of $t_1$. Where is it wrong?






sequences-and-series







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$endgroup$












  • $begingroup$
    I've never seen this approach to finding the nth term of a sequence.. I'm curious to see what others have to say.
    $endgroup$
    – T. Fo
    Jan 8 at 17:15
















1












$begingroup$


My attempt:




Let,$$space space space space space space space S=3+7+12+18+......+t_n$$
$$space space space space space space space spacespace space space space space space space space space S=space space space space space space space space 3+7+12+18+...t_{n-1}+t_n$$




Subtracting,
$$0=3+4+5+6+...(t_n-t_{n-1})-t_n$$
$$implies t_n=frac{n(n+1)}{2}-3$$
By adding and subtracting $1,2$.
However this gives a negative value of $t_1$. Where is it wrong?






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    I've never seen this approach to finding the nth term of a sequence.. I'm curious to see what others have to say.
    $endgroup$
    – T. Fo
    Jan 8 at 17:15














1












1








1





$begingroup$


My attempt:




Let,$$space space space space space space space S=3+7+12+18+......+t_n$$
$$space space space space space space space spacespace space space space space space space space space S=space space space space space space space space 3+7+12+18+...t_{n-1}+t_n$$




Subtracting,
$$0=3+4+5+6+...(t_n-t_{n-1})-t_n$$
$$implies t_n=frac{n(n+1)}{2}-3$$
By adding and subtracting $1,2$.
However this gives a negative value of $t_1$. Where is it wrong?






sequences-and-series







share|cite|improve this question









$endgroup$




My attempt:




Let,$$space space space space space space space S=3+7+12+18+......+t_n$$
$$space space space space space space space spacespace space space space space space space space space S=space space space space space space space space 3+7+12+18+...t_{n-1}+t_n$$




Subtracting,
$$0=3+4+5+6+...(t_n-t_{n-1})-t_n$$
$$implies t_n=frac{n(n+1)}{2}-3$$
By adding and subtracting $1,2$.
However this gives a negative value of $t_1$. Where is it wrong?






sequences-and-series




sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 17:12









tatantatan

5,63362655




5,63362655












  • $begingroup$
    I've never seen this approach to finding the nth term of a sequence.. I'm curious to see what others have to say.
    $endgroup$
    – T. Fo
    Jan 8 at 17:15


















  • $begingroup$
    I've never seen this approach to finding the nth term of a sequence.. I'm curious to see what others have to say.
    $endgroup$
    – T. Fo
    Jan 8 at 17:15
















$begingroup$
I've never seen this approach to finding the nth term of a sequence.. I'm curious to see what others have to say.
$endgroup$
– T. Fo
Jan 8 at 17:15




$begingroup$
I've never seen this approach to finding the nth term of a sequence.. I'm curious to see what others have to say.
$endgroup$
– T. Fo
Jan 8 at 17:15










5 Answers
5






active

oldest

votes


















2












$begingroup$

$dfrac{n^2+n}2-3=3iff0=(n+4)(n-3)$



So, your sequence start with $n=3$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Suppose I have to find the sum $S$ as given in the question. How do I do it then?
    $endgroup$
    – tatan
    Jan 8 at 17:21










  • $begingroup$
    @tatan, Use brilliant.org/wiki/sum-of-n-n2-or-n3
    $endgroup$
    – lab bhattacharjee
    Jan 8 at 17:23






  • 1




    $begingroup$
    @tatan: you did just fine, but note that $3$ is then the third term of your sum because you put $1,2$ in front. Then the last term is $t_{n+2}$. If you want to start counting your terms at $1$, you must update your formula to be $frac {(n+2)(n+3)}2-3$, which works fine.
    $endgroup$
    – Ross Millikan
    Jan 8 at 17:28










  • $begingroup$
    @RossMillikan Ahhh.... Thanks a lot ;-)
    $endgroup$
    – tatan
    Jan 8 at 17:29



















1












$begingroup$

$S= 3+7+12+18+...;$



$S= 3 +(3+4) +( 3+4+5) +(3+4+5+6) +...;$



$S= a_1 +a_2 +a_3 +a_4.......+ a_n;$



$a_n =(3+4+5+ ...(n+2));$



$a_n = 2n+(1+2+3...+n)$, or



$a_n= 2n+n(n+1)/2.$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The sum of $n$ consecutive integers, from $3$ and up to $t_n- t_{n-1}$, isn't $frac{n(n+1)}2$. If you think about which (and how many) integers $frac{n(n+1)}2$ usually represents the sum of, you will hopefully see what went wrong.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      i did make it $(1+2)+3+...(t_n-t_{n-1})-(1+2)$
      $endgroup$
      – tatan
      Jan 8 at 17:20












    • $begingroup$
      @tatan I know. I'm trying to tell you that that's not $frac{n(n+1)}2-3$.
      $endgroup$
      – Arthur
      Jan 8 at 18:38





















    0












    $begingroup$

    The systematic way is to use Newton's interpolation formula based on
    repeated differences:
    $$
    begin{array}{llll}
    3 & 7 & 12 & 18 & \
    4 & 5 & 6 & \
    1 & 1 & \
    0 & \
    end{array}
    $$
    and so
    $$
    t_n=3 binom{n-1}{0} + 4 binom{n-1}{1} + 1 binom{n-1}{2}
    $$
    which simplifies to
    $$
    t_n=dfrac{n (n + 5)}{2}
    $$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You can prove by induction that $a_{n}=3n+cfrac{n(n-1)}{2}$

      Here, you don't have to assume that $n>2$






      share|cite|improve this answer











      $endgroup$













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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        $dfrac{n^2+n}2-3=3iff0=(n+4)(n-3)$



        So, your sequence start with $n=3$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Suppose I have to find the sum $S$ as given in the question. How do I do it then?
          $endgroup$
          – tatan
          Jan 8 at 17:21










        • $begingroup$
          @tatan, Use brilliant.org/wiki/sum-of-n-n2-or-n3
          $endgroup$
          – lab bhattacharjee
          Jan 8 at 17:23






        • 1




          $begingroup$
          @tatan: you did just fine, but note that $3$ is then the third term of your sum because you put $1,2$ in front. Then the last term is $t_{n+2}$. If you want to start counting your terms at $1$, you must update your formula to be $frac {(n+2)(n+3)}2-3$, which works fine.
          $endgroup$
          – Ross Millikan
          Jan 8 at 17:28










        • $begingroup$
          @RossMillikan Ahhh.... Thanks a lot ;-)
          $endgroup$
          – tatan
          Jan 8 at 17:29
















        2












        $begingroup$

        $dfrac{n^2+n}2-3=3iff0=(n+4)(n-3)$



        So, your sequence start with $n=3$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Suppose I have to find the sum $S$ as given in the question. How do I do it then?
          $endgroup$
          – tatan
          Jan 8 at 17:21










        • $begingroup$
          @tatan, Use brilliant.org/wiki/sum-of-n-n2-or-n3
          $endgroup$
          – lab bhattacharjee
          Jan 8 at 17:23






        • 1




          $begingroup$
          @tatan: you did just fine, but note that $3$ is then the third term of your sum because you put $1,2$ in front. Then the last term is $t_{n+2}$. If you want to start counting your terms at $1$, you must update your formula to be $frac {(n+2)(n+3)}2-3$, which works fine.
          $endgroup$
          – Ross Millikan
          Jan 8 at 17:28










        • $begingroup$
          @RossMillikan Ahhh.... Thanks a lot ;-)
          $endgroup$
          – tatan
          Jan 8 at 17:29














        2












        2








        2





        $begingroup$

        $dfrac{n^2+n}2-3=3iff0=(n+4)(n-3)$



        So, your sequence start with $n=3$






        share|cite|improve this answer









        $endgroup$



        $dfrac{n^2+n}2-3=3iff0=(n+4)(n-3)$



        So, your sequence start with $n=3$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 17:16









        lab bhattacharjeelab bhattacharjee

        225k15156274




        225k15156274












        • $begingroup$
          Suppose I have to find the sum $S$ as given in the question. How do I do it then?
          $endgroup$
          – tatan
          Jan 8 at 17:21










        • $begingroup$
          @tatan, Use brilliant.org/wiki/sum-of-n-n2-or-n3
          $endgroup$
          – lab bhattacharjee
          Jan 8 at 17:23






        • 1




          $begingroup$
          @tatan: you did just fine, but note that $3$ is then the third term of your sum because you put $1,2$ in front. Then the last term is $t_{n+2}$. If you want to start counting your terms at $1$, you must update your formula to be $frac {(n+2)(n+3)}2-3$, which works fine.
          $endgroup$
          – Ross Millikan
          Jan 8 at 17:28










        • $begingroup$
          @RossMillikan Ahhh.... Thanks a lot ;-)
          $endgroup$
          – tatan
          Jan 8 at 17:29


















        • $begingroup$
          Suppose I have to find the sum $S$ as given in the question. How do I do it then?
          $endgroup$
          – tatan
          Jan 8 at 17:21










        • $begingroup$
          @tatan, Use brilliant.org/wiki/sum-of-n-n2-or-n3
          $endgroup$
          – lab bhattacharjee
          Jan 8 at 17:23






        • 1




          $begingroup$
          @tatan: you did just fine, but note that $3$ is then the third term of your sum because you put $1,2$ in front. Then the last term is $t_{n+2}$. If you want to start counting your terms at $1$, you must update your formula to be $frac {(n+2)(n+3)}2-3$, which works fine.
          $endgroup$
          – Ross Millikan
          Jan 8 at 17:28










        • $begingroup$
          @RossMillikan Ahhh.... Thanks a lot ;-)
          $endgroup$
          – tatan
          Jan 8 at 17:29
















        $begingroup$
        Suppose I have to find the sum $S$ as given in the question. How do I do it then?
        $endgroup$
        – tatan
        Jan 8 at 17:21




        $begingroup$
        Suppose I have to find the sum $S$ as given in the question. How do I do it then?
        $endgroup$
        – tatan
        Jan 8 at 17:21












        $begingroup$
        @tatan, Use brilliant.org/wiki/sum-of-n-n2-or-n3
        $endgroup$
        – lab bhattacharjee
        Jan 8 at 17:23




        $begingroup$
        @tatan, Use brilliant.org/wiki/sum-of-n-n2-or-n3
        $endgroup$
        – lab bhattacharjee
        Jan 8 at 17:23




        1




        1




        $begingroup$
        @tatan: you did just fine, but note that $3$ is then the third term of your sum because you put $1,2$ in front. Then the last term is $t_{n+2}$. If you want to start counting your terms at $1$, you must update your formula to be $frac {(n+2)(n+3)}2-3$, which works fine.
        $endgroup$
        – Ross Millikan
        Jan 8 at 17:28




        $begingroup$
        @tatan: you did just fine, but note that $3$ is then the third term of your sum because you put $1,2$ in front. Then the last term is $t_{n+2}$. If you want to start counting your terms at $1$, you must update your formula to be $frac {(n+2)(n+3)}2-3$, which works fine.
        $endgroup$
        – Ross Millikan
        Jan 8 at 17:28












        $begingroup$
        @RossMillikan Ahhh.... Thanks a lot ;-)
        $endgroup$
        – tatan
        Jan 8 at 17:29




        $begingroup$
        @RossMillikan Ahhh.... Thanks a lot ;-)
        $endgroup$
        – tatan
        Jan 8 at 17:29











        1












        $begingroup$

        $S= 3+7+12+18+...;$



        $S= 3 +(3+4) +( 3+4+5) +(3+4+5+6) +...;$



        $S= a_1 +a_2 +a_3 +a_4.......+ a_n;$



        $a_n =(3+4+5+ ...(n+2));$



        $a_n = 2n+(1+2+3...+n)$, or



        $a_n= 2n+n(n+1)/2.$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          $S= 3+7+12+18+...;$



          $S= 3 +(3+4) +( 3+4+5) +(3+4+5+6) +...;$



          $S= a_1 +a_2 +a_3 +a_4.......+ a_n;$



          $a_n =(3+4+5+ ...(n+2));$



          $a_n = 2n+(1+2+3...+n)$, or



          $a_n= 2n+n(n+1)/2.$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            $S= 3+7+12+18+...;$



            $S= 3 +(3+4) +( 3+4+5) +(3+4+5+6) +...;$



            $S= a_1 +a_2 +a_3 +a_4.......+ a_n;$



            $a_n =(3+4+5+ ...(n+2));$



            $a_n = 2n+(1+2+3...+n)$, or



            $a_n= 2n+n(n+1)/2.$






            share|cite|improve this answer









            $endgroup$



            $S= 3+7+12+18+...;$



            $S= 3 +(3+4) +( 3+4+5) +(3+4+5+6) +...;$



            $S= a_1 +a_2 +a_3 +a_4.......+ a_n;$



            $a_n =(3+4+5+ ...(n+2));$



            $a_n = 2n+(1+2+3...+n)$, or



            $a_n= 2n+n(n+1)/2.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 8 at 18:05









            Peter SzilasPeter Szilas

            11.1k2821




            11.1k2821























                0












                $begingroup$

                The sum of $n$ consecutive integers, from $3$ and up to $t_n- t_{n-1}$, isn't $frac{n(n+1)}2$. If you think about which (and how many) integers $frac{n(n+1)}2$ usually represents the sum of, you will hopefully see what went wrong.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  i did make it $(1+2)+3+...(t_n-t_{n-1})-(1+2)$
                  $endgroup$
                  – tatan
                  Jan 8 at 17:20












                • $begingroup$
                  @tatan I know. I'm trying to tell you that that's not $frac{n(n+1)}2-3$.
                  $endgroup$
                  – Arthur
                  Jan 8 at 18:38


















                0












                $begingroup$

                The sum of $n$ consecutive integers, from $3$ and up to $t_n- t_{n-1}$, isn't $frac{n(n+1)}2$. If you think about which (and how many) integers $frac{n(n+1)}2$ usually represents the sum of, you will hopefully see what went wrong.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  i did make it $(1+2)+3+...(t_n-t_{n-1})-(1+2)$
                  $endgroup$
                  – tatan
                  Jan 8 at 17:20












                • $begingroup$
                  @tatan I know. I'm trying to tell you that that's not $frac{n(n+1)}2-3$.
                  $endgroup$
                  – Arthur
                  Jan 8 at 18:38
















                0












                0








                0





                $begingroup$

                The sum of $n$ consecutive integers, from $3$ and up to $t_n- t_{n-1}$, isn't $frac{n(n+1)}2$. If you think about which (and how many) integers $frac{n(n+1)}2$ usually represents the sum of, you will hopefully see what went wrong.






                share|cite|improve this answer









                $endgroup$



                The sum of $n$ consecutive integers, from $3$ and up to $t_n- t_{n-1}$, isn't $frac{n(n+1)}2$. If you think about which (and how many) integers $frac{n(n+1)}2$ usually represents the sum of, you will hopefully see what went wrong.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 17:19









                ArthurArthur

                113k7110193




                113k7110193












                • $begingroup$
                  i did make it $(1+2)+3+...(t_n-t_{n-1})-(1+2)$
                  $endgroup$
                  – tatan
                  Jan 8 at 17:20












                • $begingroup$
                  @tatan I know. I'm trying to tell you that that's not $frac{n(n+1)}2-3$.
                  $endgroup$
                  – Arthur
                  Jan 8 at 18:38




















                • $begingroup$
                  i did make it $(1+2)+3+...(t_n-t_{n-1})-(1+2)$
                  $endgroup$
                  – tatan
                  Jan 8 at 17:20












                • $begingroup$
                  @tatan I know. I'm trying to tell you that that's not $frac{n(n+1)}2-3$.
                  $endgroup$
                  – Arthur
                  Jan 8 at 18:38


















                $begingroup$
                i did make it $(1+2)+3+...(t_n-t_{n-1})-(1+2)$
                $endgroup$
                – tatan
                Jan 8 at 17:20






                $begingroup$
                i did make it $(1+2)+3+...(t_n-t_{n-1})-(1+2)$
                $endgroup$
                – tatan
                Jan 8 at 17:20














                $begingroup$
                @tatan I know. I'm trying to tell you that that's not $frac{n(n+1)}2-3$.
                $endgroup$
                – Arthur
                Jan 8 at 18:38






                $begingroup$
                @tatan I know. I'm trying to tell you that that's not $frac{n(n+1)}2-3$.
                $endgroup$
                – Arthur
                Jan 8 at 18:38













                0












                $begingroup$

                The systematic way is to use Newton's interpolation formula based on
                repeated differences:
                $$
                begin{array}{llll}
                3 & 7 & 12 & 18 & \
                4 & 5 & 6 & \
                1 & 1 & \
                0 & \
                end{array}
                $$
                and so
                $$
                t_n=3 binom{n-1}{0} + 4 binom{n-1}{1} + 1 binom{n-1}{2}
                $$
                which simplifies to
                $$
                t_n=dfrac{n (n + 5)}{2}
                $$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  The systematic way is to use Newton's interpolation formula based on
                  repeated differences:
                  $$
                  begin{array}{llll}
                  3 & 7 & 12 & 18 & \
                  4 & 5 & 6 & \
                  1 & 1 & \
                  0 & \
                  end{array}
                  $$
                  and so
                  $$
                  t_n=3 binom{n-1}{0} + 4 binom{n-1}{1} + 1 binom{n-1}{2}
                  $$
                  which simplifies to
                  $$
                  t_n=dfrac{n (n + 5)}{2}
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    The systematic way is to use Newton's interpolation formula based on
                    repeated differences:
                    $$
                    begin{array}{llll}
                    3 & 7 & 12 & 18 & \
                    4 & 5 & 6 & \
                    1 & 1 & \
                    0 & \
                    end{array}
                    $$
                    and so
                    $$
                    t_n=3 binom{n-1}{0} + 4 binom{n-1}{1} + 1 binom{n-1}{2}
                    $$
                    which simplifies to
                    $$
                    t_n=dfrac{n (n + 5)}{2}
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    The systematic way is to use Newton's interpolation formula based on
                    repeated differences:
                    $$
                    begin{array}{llll}
                    3 & 7 & 12 & 18 & \
                    4 & 5 & 6 & \
                    1 & 1 & \
                    0 & \
                    end{array}
                    $$
                    and so
                    $$
                    t_n=3 binom{n-1}{0} + 4 binom{n-1}{1} + 1 binom{n-1}{2}
                    $$
                    which simplifies to
                    $$
                    t_n=dfrac{n (n + 5)}{2}
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 8 at 17:25









                    lhflhf

                    164k10170395




                    164k10170395























                        0












                        $begingroup$

                        You can prove by induction that $a_{n}=3n+cfrac{n(n-1)}{2}$

                        Here, you don't have to assume that $n>2$






                        share|cite|improve this answer











                        $endgroup$


















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                          $begingroup$

                          You can prove by induction that $a_{n}=3n+cfrac{n(n-1)}{2}$

                          Here, you don't have to assume that $n>2$






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You can prove by induction that $a_{n}=3n+cfrac{n(n-1)}{2}$

                            Here, you don't have to assume that $n>2$






                            share|cite|improve this answer











                            $endgroup$



                            You can prove by induction that $a_{n}=3n+cfrac{n(n-1)}{2}$

                            Here, you don't have to assume that $n>2$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 8 at 17:58

























                            answered Jan 8 at 17:19









                            The CatThe Cat

                            18611




                            18611






























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