Finding sum of none arithmetic series












1












$begingroup$


I have a question to find the sum of the following sum:
$$
S = small{1*1+2*3+3*5+4*7+...+100*199}
$$
I figured out that for each element in this series the following holds:
$$
a_n = a_{n-1} + 4n - 3
$$
But I don't know where to go from here, I tried subtracting some other series but that did not work very well






sequences-and-series summation telescopic-series







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    1












    $begingroup$


    I have a question to find the sum of the following sum:
    $$
    S = small{1*1+2*3+3*5+4*7+...+100*199}
    $$
    I figured out that for each element in this series the following holds:
    $$
    a_n = a_{n-1} + 4n - 3
    $$
    But I don't know where to go from here, I tried subtracting some other series but that did not work very well






    sequences-and-series summation telescopic-series







    share|cite|improve this question











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      1








      1


      0



      $begingroup$


      I have a question to find the sum of the following sum:
      $$
      S = small{1*1+2*3+3*5+4*7+...+100*199}
      $$
      I figured out that for each element in this series the following holds:
      $$
      a_n = a_{n-1} + 4n - 3
      $$
      But I don't know where to go from here, I tried subtracting some other series but that did not work very well






      sequences-and-series summation telescopic-series







      share|cite|improve this question











      $endgroup$




      I have a question to find the sum of the following sum:
      $$
      S = small{1*1+2*3+3*5+4*7+...+100*199}
      $$
      I figured out that for each element in this series the following holds:
      $$
      a_n = a_{n-1} + 4n - 3
      $$
      But I don't know where to go from here, I tried subtracting some other series but that did not work very well






      sequences-and-series summation telescopic-series




      sequences-and-series summation telescopic-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 19 mins ago









      Michael Rozenberg

      99.4k1590189




      99.4k1590189










      asked 1 hour ago









      Guysudai1Guysudai1

      1149




      1149






















          4 Answers
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          $begingroup$

          Each term in the equation is $n(2n-1)$, so $$S=sum_{i=1}^{100}{n(2n-1)}=2sum_{i=1}^{100} n^2-sum_{i=1}^{100} n=2frac{m(m+1)(2m+1)}{6}-frac{m(m+1)}{2}$$
          where $m=100$.






          share|cite|improve this answer









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            1












            $begingroup$

            This is, for $n=100$,
            $s(n)
            =sum_{k=1}^n k(2k-1)
            =sum_{k=1}^n (2k^2-k)
            =2sum_{k=1}^n k^2
            -sum_{k=1}^n k
            $            .



            Plug in the formulas for the sums
            and you are done.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              $a_n=sum_{r=1}^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



              $$sum_{n=1}^ma_n=2sum_{n=1}^mn^2-sum_{n=1}^mn+a_0sum_{n=1}^m1$$



              Here $a_0=0$



              Alternatively,



              $$a_m=b_m+a+bm+cm^2$$



              $$4n-3=a_n-a_{n-1}=b_n-b_{n-1}+b+c(2n-1)=b_n-b_{n-1}+2c(n)+b-c$$



              WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_{n-1}$



              set $a=0$ so that $b_m=a_m=0$






              share|cite|improve this answer











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              • $begingroup$
                Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                $endgroup$
                – lab bhattacharjee
                1 hour ago



















              1












              $begingroup$

              By the telescoping sum we obtain: $$sum_{n=1}^{100}n(2n-1)=sum_{n=1}^{100}left(frac{n(n+1)(4n-1)}{6}-frac{(n-1)n(4n-5)}{6}right)=$$
              $$=frac{100cdot101cdot399}{6}=671650.$$






              share|cite|improve this answer









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                4 Answers
                4






                active

                oldest

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                4 Answers
                4






                active

                oldest

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                active

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                active

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                2












                $begingroup$

                Each term in the equation is $n(2n-1)$, so $$S=sum_{i=1}^{100}{n(2n-1)}=2sum_{i=1}^{100} n^2-sum_{i=1}^{100} n=2frac{m(m+1)(2m+1)}{6}-frac{m(m+1)}{2}$$
                where $m=100$.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Each term in the equation is $n(2n-1)$, so $$S=sum_{i=1}^{100}{n(2n-1)}=2sum_{i=1}^{100} n^2-sum_{i=1}^{100} n=2frac{m(m+1)(2m+1)}{6}-frac{m(m+1)}{2}$$
                  where $m=100$.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Each term in the equation is $n(2n-1)$, so $$S=sum_{i=1}^{100}{n(2n-1)}=2sum_{i=1}^{100} n^2-sum_{i=1}^{100} n=2frac{m(m+1)(2m+1)}{6}-frac{m(m+1)}{2}$$
                    where $m=100$.






                    share|cite|improve this answer









                    $endgroup$



                    Each term in the equation is $n(2n-1)$, so $$S=sum_{i=1}^{100}{n(2n-1)}=2sum_{i=1}^{100} n^2-sum_{i=1}^{100} n=2frac{m(m+1)(2m+1)}{6}-frac{m(m+1)}{2}$$
                    where $m=100$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    gunesgunes

                    2645




                    2645























                        1












                        $begingroup$

                        This is, for $n=100$,
                        $s(n)
                        =sum_{k=1}^n k(2k-1)
                        =sum_{k=1}^n (2k^2-k)
                        =2sum_{k=1}^n k^2
                        -sum_{k=1}^n k
                        $                        .



                        Plug in the formulas for the sums
                        and you are done.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          This is, for $n=100$,
                          $s(n)
                          =sum_{k=1}^n k(2k-1)
                          =sum_{k=1}^n (2k^2-k)
                          =2sum_{k=1}^n k^2
                          -sum_{k=1}^n k
                          $                          .



                          Plug in the formulas for the sums
                          and you are done.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            This is, for $n=100$,
                            $s(n)
                            =sum_{k=1}^n k(2k-1)
                            =sum_{k=1}^n (2k^2-k)
                            =2sum_{k=1}^n k^2
                            -sum_{k=1}^n k
                            $                            .



                            Plug in the formulas for the sums
                            and you are done.






                            share|cite|improve this answer









                            $endgroup$



                            This is, for $n=100$,
                            $s(n)
                            =sum_{k=1}^n k(2k-1)
                            =sum_{k=1}^n (2k^2-k)
                            =2sum_{k=1}^n k^2
                            -sum_{k=1}^n k
                            $                            .



                            Plug in the formulas for the sums
                            and you are done.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            marty cohenmarty cohen

                            73.4k549128




                            73.4k549128























                                1












                                $begingroup$

                                $a_n=sum_{r=1}^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



                                $$sum_{n=1}^ma_n=2sum_{n=1}^mn^2-sum_{n=1}^mn+a_0sum_{n=1}^m1$$



                                Here $a_0=0$



                                Alternatively,



                                $$a_m=b_m+a+bm+cm^2$$



                                $$4n-3=a_n-a_{n-1}=b_n-b_{n-1}+b+c(2n-1)=b_n-b_{n-1}+2c(n)+b-c$$



                                WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_{n-1}$



                                set $a=0$ so that $b_m=a_m=0$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                  $endgroup$
                                  – lab bhattacharjee
                                  1 hour ago
















                                1












                                $begingroup$

                                $a_n=sum_{r=1}^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



                                $$sum_{n=1}^ma_n=2sum_{n=1}^mn^2-sum_{n=1}^mn+a_0sum_{n=1}^m1$$



                                Here $a_0=0$



                                Alternatively,



                                $$a_m=b_m+a+bm+cm^2$$



                                $$4n-3=a_n-a_{n-1}=b_n-b_{n-1}+b+c(2n-1)=b_n-b_{n-1}+2c(n)+b-c$$



                                WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_{n-1}$



                                set $a=0$ so that $b_m=a_m=0$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                  $endgroup$
                                  – lab bhattacharjee
                                  1 hour ago














                                1












                                1








                                1





                                $begingroup$

                                $a_n=sum_{r=1}^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



                                $$sum_{n=1}^ma_n=2sum_{n=1}^mn^2-sum_{n=1}^mn+a_0sum_{n=1}^m1$$



                                Here $a_0=0$



                                Alternatively,



                                $$a_m=b_m+a+bm+cm^2$$



                                $$4n-3=a_n-a_{n-1}=b_n-b_{n-1}+b+c(2n-1)=b_n-b_{n-1}+2c(n)+b-c$$



                                WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_{n-1}$



                                set $a=0$ so that $b_m=a_m=0$






                                share|cite|improve this answer











                                $endgroup$



                                $a_n=sum_{r=1}^n(4r-3)+a_0=dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$



                                $$sum_{n=1}^ma_n=2sum_{n=1}^mn^2-sum_{n=1}^mn+a_0sum_{n=1}^m1$$



                                Here $a_0=0$



                                Alternatively,



                                $$a_m=b_m+a+bm+cm^2$$



                                $$4n-3=a_n-a_{n-1}=b_n-b_{n-1}+b+c(2n-1)=b_n-b_{n-1}+2c(n)+b-c$$



                                WLOG set $2c=4,b-c=-3iff c=b+3$ to find $b_n=b_{n-1}$



                                set $a=0$ so that $b_m=a_m=0$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited 1 hour ago

























                                answered 1 hour ago









                                lab bhattacharjeelab bhattacharjee

                                225k15156274




                                225k15156274












                                • $begingroup$
                                  Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                  $endgroup$
                                  – lab bhattacharjee
                                  1 hour ago


















                                • $begingroup$
                                  Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                  $endgroup$
                                  – lab bhattacharjee
                                  1 hour ago
















                                $begingroup$
                                Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                $endgroup$
                                – lab bhattacharjee
                                1 hour ago




                                $begingroup$
                                Use math.stackexchange.com/questions/1806906/… or math.stackexchange.com/questions/2992733/… or math.stackexchange.com/questions/1339732/… or math.stackexchange.com/questions/183316/…
                                $endgroup$
                                – lab bhattacharjee
                                1 hour ago











                                1












                                $begingroup$

                                By the telescoping sum we obtain: $$sum_{n=1}^{100}n(2n-1)=sum_{n=1}^{100}left(frac{n(n+1)(4n-1)}{6}-frac{(n-1)n(4n-5)}{6}right)=$$
                                $$=frac{100cdot101cdot399}{6}=671650.$$






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  By the telescoping sum we obtain: $$sum_{n=1}^{100}n(2n-1)=sum_{n=1}^{100}left(frac{n(n+1)(4n-1)}{6}-frac{(n-1)n(4n-5)}{6}right)=$$
                                  $$=frac{100cdot101cdot399}{6}=671650.$$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    By the telescoping sum we obtain: $$sum_{n=1}^{100}n(2n-1)=sum_{n=1}^{100}left(frac{n(n+1)(4n-1)}{6}-frac{(n-1)n(4n-5)}{6}right)=$$
                                    $$=frac{100cdot101cdot399}{6}=671650.$$






                                    share|cite|improve this answer









                                    $endgroup$



                                    By the telescoping sum we obtain: $$sum_{n=1}^{100}n(2n-1)=sum_{n=1}^{100}left(frac{n(n+1)(4n-1)}{6}-frac{(n-1)n(4n-5)}{6}right)=$$
                                    $$=frac{100cdot101cdot399}{6}=671650.$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 31 mins ago









                                    Michael RozenbergMichael Rozenberg

                                    99.4k1590189




                                    99.4k1590189






























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