What would be the closed form of $ncdot n + (n-2)cdot(n-2) + (n-4)cdot(n-4) + dots$?












0














As part of solving this problem , I came up with the following expression:



$$text{ans} = ncdot n + (n-2)cdot (n-2) + (n-4)cdot (n-4) + dots$$



So I am just running a loop to calculate the ans. However, since there is a pattern in the form of a series, I am wondering if a closed form exists for it? Please also show the steps to obtain so that I can learn, in general, how to approach such problems for finding closed forms.



EDIT: series goes till 2 if n is even, and till 1 if n is odd.






sequences-and-series closed-form







share|cite|improve this question
























  • Do you know the formula for the sum of the first $m$ squares?
    – Arnaud D.
    Aug 17 '18 at 9:55










  • yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
    – user3243499
    Aug 17 '18 at 10:00










  • It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
    – Claude Leibovici
    Aug 17 '18 at 10:03










  • Where exactly does it stop?
    – Arnaud Mortier
    Aug 17 '18 at 10:03










  • @user3243499: no, it's the sum of the squares of the same parity.
    – Yves Daoust
    Aug 17 '18 at 10:04
















0














As part of solving this problem , I came up with the following expression:



$$text{ans} = ncdot n + (n-2)cdot (n-2) + (n-4)cdot (n-4) + dots$$



So I am just running a loop to calculate the ans. However, since there is a pattern in the form of a series, I am wondering if a closed form exists for it? Please also show the steps to obtain so that I can learn, in general, how to approach such problems for finding closed forms.



EDIT: series goes till 2 if n is even, and till 1 if n is odd.






sequences-and-series closed-form







share|cite|improve this question
























  • Do you know the formula for the sum of the first $m$ squares?
    – Arnaud D.
    Aug 17 '18 at 9:55










  • yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
    – user3243499
    Aug 17 '18 at 10:00










  • It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
    – Claude Leibovici
    Aug 17 '18 at 10:03










  • Where exactly does it stop?
    – Arnaud Mortier
    Aug 17 '18 at 10:03










  • @user3243499: no, it's the sum of the squares of the same parity.
    – Yves Daoust
    Aug 17 '18 at 10:04














0












0








0







As part of solving this problem , I came up with the following expression:



$$text{ans} = ncdot n + (n-2)cdot (n-2) + (n-4)cdot (n-4) + dots$$



So I am just running a loop to calculate the ans. However, since there is a pattern in the form of a series, I am wondering if a closed form exists for it? Please also show the steps to obtain so that I can learn, in general, how to approach such problems for finding closed forms.



EDIT: series goes till 2 if n is even, and till 1 if n is odd.






sequences-and-series closed-form







share|cite|improve this question















As part of solving this problem , I came up with the following expression:



$$text{ans} = ncdot n + (n-2)cdot (n-2) + (n-4)cdot (n-4) + dots$$



So I am just running a loop to calculate the ans. However, since there is a pattern in the form of a series, I am wondering if a closed form exists for it? Please also show the steps to obtain so that I can learn, in general, how to approach such problems for finding closed forms.



EDIT: series goes till 2 if n is even, and till 1 if n is odd.






sequences-and-series closed-form




sequences-and-series closed-form






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 17 '18 at 10:12

























asked Aug 17 '18 at 9:54









user3243499

192113




192113












  • Do you know the formula for the sum of the first $m$ squares?
    – Arnaud D.
    Aug 17 '18 at 9:55










  • yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
    – user3243499
    Aug 17 '18 at 10:00










  • It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
    – Claude Leibovici
    Aug 17 '18 at 10:03










  • Where exactly does it stop?
    – Arnaud Mortier
    Aug 17 '18 at 10:03










  • @user3243499: no, it's the sum of the squares of the same parity.
    – Yves Daoust
    Aug 17 '18 at 10:04


















  • Do you know the formula for the sum of the first $m$ squares?
    – Arnaud D.
    Aug 17 '18 at 9:55










  • yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
    – user3243499
    Aug 17 '18 at 10:00










  • It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
    – Claude Leibovici
    Aug 17 '18 at 10:03










  • Where exactly does it stop?
    – Arnaud Mortier
    Aug 17 '18 at 10:03










  • @user3243499: no, it's the sum of the squares of the same parity.
    – Yves Daoust
    Aug 17 '18 at 10:04
















Do you know the formula for the sum of the first $m$ squares?
– Arnaud D.
Aug 17 '18 at 9:55




Do you know the formula for the sum of the first $m$ squares?
– Arnaud D.
Aug 17 '18 at 9:55












yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
– user3243499
Aug 17 '18 at 10:00




yes, n(n+1)(2n+1)/6. Ok, I got it. This is exactly a standard square sum sequences. How stupid I am. Thanks.
– user3243499
Aug 17 '18 at 10:00












It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
– Claude Leibovici
Aug 17 '18 at 10:03




It would be more amazing to compute $sum_{k=0}^p (n-2k)^2$ for $pleq frac n2$. The formula is nice
– Claude Leibovici
Aug 17 '18 at 10:03












Where exactly does it stop?
– Arnaud Mortier
Aug 17 '18 at 10:03




Where exactly does it stop?
– Arnaud Mortier
Aug 17 '18 at 10:03












@user3243499: no, it's the sum of the squares of the same parity.
– Yves Daoust
Aug 17 '18 at 10:04




@user3243499: no, it's the sum of the squares of the same parity.
– Yves Daoust
Aug 17 '18 at 10:04










5 Answers
5






active

oldest

votes


















3














It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.



So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.



For even $n$,



$$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$



For odd $n$,



$$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$






share|cite|improve this answer





























    2














    If $n$ is even, you have



    $
    begin{align}
    sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
    &=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
    &=frac{n(n+1)(n+2)}6 \
    text{If $n$ is odd, you have} \
    \
    sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
    &=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
    &=frac{n(n+1)(n+2)}6
    end{align}$






    share|cite|improve this answer





























      1














      Use the classic formula for the sum of increasing squares and some manipulation to find:



      $$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
      frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
      +3right)$$






      share|cite|improve this answer





















      • I don't think that the formula is right for the even $n$.
        – Yves Daoust
        Aug 17 '18 at 10:17



















      0














      An alternative method:



      Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is



      $0,1,4,10,20,35, 56$



      The differences between these terms are:



      $1,3,6,10,15,21$



      which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that



      $s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$



      $= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$



      $=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$



      $=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$



      $ = frac{n(n+1)(n+2)}{6}$



      This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction






      share|cite|improve this answer





























        0














        Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$






        share|cite|improve this answer





















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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.



          So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.



          For even $n$,



          $$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$



          For odd $n$,



          $$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$






          share|cite|improve this answer


























            3














            It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.



            So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.



            For even $n$,



            $$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$



            For odd $n$,



            $$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$






            share|cite|improve this answer
























              3












              3








              3






              It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.



              So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.



              For even $n$,



              $$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$



              For odd $n$,



              $$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$






              share|cite|improve this answer












              It is well known that the sum of terms that are a polynomial expression of degree $d$, evaluated at the integers, is a polynomial expression of degree $d+1$.



              So if you consider four values of the sum, you can obtain the requested expression as the Lagrangian interpolation polynomial on these four points.



              For even $n$,



              $$(0,0),(2,4),(4,20),(6,56)tofrac{n^3+3n^2+2n}6.$$



              For odd $n$,



              $$(1,1),(3,10),(5,35),(7,84)tofrac{n^3+3n^2+2n}6.$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 17 '18 at 10:11









              Yves Daoust

              124k671222




              124k671222























                  2














                  If $n$ is even, you have



                  $
                  begin{align}
                  sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
                  &=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
                  &=frac{n(n+1)(n+2)}6 \
                  text{If $n$ is odd, you have} \
                  \
                  sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
                  &=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
                  &=frac{n(n+1)(n+2)}6
                  end{align}$






                  share|cite|improve this answer


























                    2














                    If $n$ is even, you have



                    $
                    begin{align}
                    sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
                    &=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
                    &=frac{n(n+1)(n+2)}6 \
                    text{If $n$ is odd, you have} \
                    \
                    sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
                    &=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
                    &=frac{n(n+1)(n+2)}6
                    end{align}$






                    share|cite|improve this answer
























                      2












                      2








                      2






                      If $n$ is even, you have



                      $
                      begin{align}
                      sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
                      &=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
                      &=frac{n(n+1)(n+2)}6 \
                      text{If $n$ is odd, you have} \
                      \
                      sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
                      &=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
                      &=frac{n(n+1)(n+2)}6
                      end{align}$






                      share|cite|improve this answer












                      If $n$ is even, you have



                      $
                      begin{align}
                      sum_{k=1}^{frac{n}2}(2k)^2&=4sum_{k=1}^{frac{n}2}k^2\
                      &=4cdotfrac{frac{n}2(frac{n}2+1)(2cdotfrac{n}2+1)}6\
                      &=frac{n(n+1)(n+2)}6 \
                      text{If $n$ is odd, you have} \
                      \
                      sum_{k=1}^{frac{n+1}2}(2k-1)^2&=4sum_{k=1}^{frac{n+1}2}k^2-4sum_{k=1}^{frac{n+1}2}k+sum_{k=1}^{frac{n+1}2}1 \
                      &=4cdotfrac{frac{n+1}2(frac{n+1}2+1)(2cdotfrac{n+1}2+1)}6-4cdotfrac{frac{n+1}2(frac{n+1}2+1)}2+frac{n+1}2 \
                      &=frac{n(n+1)(n+2)}6
                      end{align}$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Aug 17 '18 at 11:53









                      Abraham Zhang

                      596312




                      596312























                          1














                          Use the classic formula for the sum of increasing squares and some manipulation to find:



                          $$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
                          frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
                          +3right)$$






                          share|cite|improve this answer





















                          • I don't think that the formula is right for the even $n$.
                            – Yves Daoust
                            Aug 17 '18 at 10:17
















                          1














                          Use the classic formula for the sum of increasing squares and some manipulation to find:



                          $$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
                          frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
                          +3right)$$






                          share|cite|improve this answer





















                          • I don't think that the formula is right for the even $n$.
                            – Yves Daoust
                            Aug 17 '18 at 10:17














                          1












                          1








                          1






                          Use the classic formula for the sum of increasing squares and some manipulation to find:



                          $$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
                          frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
                          +3right)$$






                          share|cite|improve this answer












                          Use the classic formula for the sum of increasing squares and some manipulation to find:



                          $$frac{1}{3} left(leftlfloor frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor
                          frac{n-1}{2}rightrfloor +1right) left(2 leftlfloor frac{n-1}{2}rightrfloor
                          +3right)$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 17 '18 at 10:04









                          David G. Stork

                          9,93521232




                          9,93521232












                          • I don't think that the formula is right for the even $n$.
                            – Yves Daoust
                            Aug 17 '18 at 10:17


















                          • I don't think that the formula is right for the even $n$.
                            – Yves Daoust
                            Aug 17 '18 at 10:17
















                          I don't think that the formula is right for the even $n$.
                          – Yves Daoust
                          Aug 17 '18 at 10:17




                          I don't think that the formula is right for the even $n$.
                          – Yves Daoust
                          Aug 17 '18 at 10:17











                          0














                          An alternative method:



                          Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is



                          $0,1,4,10,20,35, 56$



                          The differences between these terms are:



                          $1,3,6,10,15,21$



                          which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that



                          $s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$



                          $= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$



                          $=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$



                          $=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$



                          $ = frac{n(n+1)(n+2)}{6}$



                          This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction






                          share|cite|improve this answer


























                            0














                            An alternative method:



                            Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is



                            $0,1,4,10,20,35, 56$



                            The differences between these terms are:



                            $1,3,6,10,15,21$



                            which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that



                            $s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$



                            $= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$



                            $=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$



                            $=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$



                            $ = frac{n(n+1)(n+2)}{6}$



                            This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction






                            share|cite|improve this answer
























                              0












                              0








                              0






                              An alternative method:



                              Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is



                              $0,1,4,10,20,35, 56$



                              The differences between these terms are:



                              $1,3,6,10,15,21$



                              which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that



                              $s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$



                              $= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$



                              $=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$



                              $=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$



                              $ = frac{n(n+1)(n+2)}{6}$



                              This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction






                              share|cite|improve this answer












                              An alternative method:



                              Let's call the sum of alternate squares up to $n$ by $s(n)$.iafor $n=0,1,2,3,4,5,6$ is



                              $0,1,4,10,20,35, 56$



                              The differences between these terms are:



                              $1,3,6,10,15,21$



                              which are the triangular numbers $frac{n(n+1)}{2}$. So this suggests that



                              $s(n) = sum_{m=0}^{n} frac{m(m+1)}{2}$



                              $= sum_{m=0}^{n} (frac{m^2}{2} + frac{m}{2})$



                              $=frac{n(n+1)(2n+1)}{12} + frac{n(n+1)}{4}$



                              $=frac{n(n+1)(2n+1) + 3n(n+1)}{12}$



                              $ = frac{n(n+1)(n+2)}{6}$



                              This is not a formal proof, but it does suggest a formula which can form the starting point of a proof by induction







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                              answered Aug 17 '18 at 11:29









                              gandalf61

                              7,718623




                              7,718623























                                  0














                                  Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$






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                                    0














                                    Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$






                                      share|cite|improve this answer












                                      Since $x^2=frac{x(x+1)}2+frac{(x-1)x}2$, your sum becomes $$sum_{k=1}^nfrac{k(k+1)}2=frac{n(n+1)(n+2)}6$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



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                                      answered Dec 28 '18 at 0:29









                                      Abraham Zhang

                                      596312




                                      596312






























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