Sum to n terms of the following series: $2 cdot 2 + 6 cdot 4 + 12 cdot 8 + 20 cdot 16 + cdots $

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I able to get the general term or $large n$th term of the series is: $$U_n=(n^2+n)2^n$$ Now i want to get $S_n$ by the method of difference.That's why I need to make $$U_n=V_r-V_r-1$$
My book is providing that $U_n=(2n^2-2n+4)2^n-[2(n-1)^2-2(n-1)+4)2^n-1]$.Now my question is how they get $U_n$ like this.Thanks in advance.




calculus summation summation-method




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    I able to get the general term or $large n$th term of the series is: $$U_n=(n^2+n)2^n$$ Now i want to get $S_n$ by the method of difference.That's why I need to make $$U_n=V_r-V_r-1$$
    My book is providing that $U_n=(2n^2-2n+4)2^n-[2(n-1)^2-2(n-1)+4)2^n-1]$.Now my question is how they get $U_n$ like this.Thanks in advance.




    calculus summation summation-method




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      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I able to get the general term or $large n$th term of the series is: $$U_n=(n^2+n)2^n$$ Now i want to get $S_n$ by the method of difference.That's why I need to make $$U_n=V_r-V_r-1$$
      My book is providing that $U_n=(2n^2-2n+4)2^n-[2(n-1)^2-2(n-1)+4)2^n-1]$.Now my question is how they get $U_n$ like this.Thanks in advance.




      calculus summation summation-method




      share|cite|improve this question














      I able to get the general term or $large n$th term of the series is: $$U_n=(n^2+n)2^n$$ Now i want to get $S_n$ by the method of difference.That's why I need to make $$U_n=V_r-V_r-1$$
      My book is providing that $U_n=(2n^2-2n+4)2^n-[2(n-1)^2-2(n-1)+4)2^n-1]$.Now my question is how they get $U_n$ like this.Thanks in advance.





      calculus summation summation-method





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      edited Aug 27 at 16:45









      Sobi

      2,418315




      2,418315










      asked Aug 27 at 16:37









      md emon

      878




      878




















          2 Answers
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          We may expect the form of $V_n$ to be $(an^2+bn+c)2^n$, since the polynomial factor of $U_n $ is in degree $2$.



          Now



          beginalign
          V_n-V_n-1&=(an^2+bn+c)2^n-[a(n-1)^2+b(n-1)+c]2^n-1\
          &=2^n-1[2an^2+2bn+2c-a(n^2-2n+1)-b(n-1)-c]\
          &=2^n-1[an^2+(2b+2a-b)n+(2c-a+b-c)]\
          &=2^n[frac a2 n^2+(a+frac b2)n+fracc+b-a2]
          endalign



          Comparing the coefficients, we get
          $$a=2,b=-2,c=4$$



          So $$V_n=(2n^2-2n+4)2^n$$






          share|cite|improve this answer




















          • Thanks a lot @Mythomorphic That's the thing I really needed :)
            – md emon
            Aug 27 at 17:39


















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          Just rearrange to obtain
          beginalign
          n^2+n &= 2n^2-n^2 -2n+3n+4-4\
          &= (2n^2-2n+4)-(n^2-3n+4)\
          &= (2n^2-2n+4) - (n^2-2n+1-n+1+2)\
          &= (2n^2-2n+4) - ((n-1)^2-(n-1)+2)\
          &= (2n^2-2n+4) - frac2(n-1)^2-2(n-1)+42.
          endalign
          Do you see it now?






          share|cite|improve this answer




















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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

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            up vote
            1
            down vote













            We may expect the form of $V_n$ to be $(an^2+bn+c)2^n$, since the polynomial factor of $U_n $ is in degree $2$.



            Now



            beginalign
            V_n-V_n-1&=(an^2+bn+c)2^n-[a(n-1)^2+b(n-1)+c]2^n-1\
            &=2^n-1[2an^2+2bn+2c-a(n^2-2n+1)-b(n-1)-c]\
            &=2^n-1[an^2+(2b+2a-b)n+(2c-a+b-c)]\
            &=2^n[frac a2 n^2+(a+frac b2)n+fracc+b-a2]
            endalign



            Comparing the coefficients, we get
            $$a=2,b=-2,c=4$$



            So $$V_n=(2n^2-2n+4)2^n$$






            share|cite|improve this answer




















            • Thanks a lot @Mythomorphic That's the thing I really needed :)
              – md emon
              Aug 27 at 17:39















            up vote
            1
            down vote













            We may expect the form of $V_n$ to be $(an^2+bn+c)2^n$, since the polynomial factor of $U_n $ is in degree $2$.



            Now



            beginalign
            V_n-V_n-1&=(an^2+bn+c)2^n-[a(n-1)^2+b(n-1)+c]2^n-1\
            &=2^n-1[2an^2+2bn+2c-a(n^2-2n+1)-b(n-1)-c]\
            &=2^n-1[an^2+(2b+2a-b)n+(2c-a+b-c)]\
            &=2^n[frac a2 n^2+(a+frac b2)n+fracc+b-a2]
            endalign



            Comparing the coefficients, we get
            $$a=2,b=-2,c=4$$



            So $$V_n=(2n^2-2n+4)2^n$$






            share|cite|improve this answer




















            • Thanks a lot @Mythomorphic That's the thing I really needed :)
              – md emon
              Aug 27 at 17:39













            up vote
            1
            down vote










            up vote
            1
            down vote









            We may expect the form of $V_n$ to be $(an^2+bn+c)2^n$, since the polynomial factor of $U_n $ is in degree $2$.



            Now



            beginalign
            V_n-V_n-1&=(an^2+bn+c)2^n-[a(n-1)^2+b(n-1)+c]2^n-1\
            &=2^n-1[2an^2+2bn+2c-a(n^2-2n+1)-b(n-1)-c]\
            &=2^n-1[an^2+(2b+2a-b)n+(2c-a+b-c)]\
            &=2^n[frac a2 n^2+(a+frac b2)n+fracc+b-a2]
            endalign



            Comparing the coefficients, we get
            $$a=2,b=-2,c=4$$



            So $$V_n=(2n^2-2n+4)2^n$$






            share|cite|improve this answer












            We may expect the form of $V_n$ to be $(an^2+bn+c)2^n$, since the polynomial factor of $U_n $ is in degree $2$.



            Now



            beginalign
            V_n-V_n-1&=(an^2+bn+c)2^n-[a(n-1)^2+b(n-1)+c]2^n-1\
            &=2^n-1[2an^2+2bn+2c-a(n^2-2n+1)-b(n-1)-c]\
            &=2^n-1[an^2+(2b+2a-b)n+(2c-a+b-c)]\
            &=2^n[frac a2 n^2+(a+frac b2)n+fracc+b-a2]
            endalign



            Comparing the coefficients, we get
            $$a=2,b=-2,c=4$$



            So $$V_n=(2n^2-2n+4)2^n$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 27 at 17:18









            Mythomorphic

            5,1911733




            5,1911733











            • Thanks a lot @Mythomorphic That's the thing I really needed :)
              – md emon
              Aug 27 at 17:39

















            • Thanks a lot @Mythomorphic That's the thing I really needed :)
              – md emon
              Aug 27 at 17:39
















            Thanks a lot @Mythomorphic That's the thing I really needed :)
            – md emon
            Aug 27 at 17:39





            Thanks a lot @Mythomorphic That's the thing I really needed :)
            – md emon
            Aug 27 at 17:39











            up vote
            0
            down vote













            Just rearrange to obtain
            beginalign
            n^2+n &= 2n^2-n^2 -2n+3n+4-4\
            &= (2n^2-2n+4)-(n^2-3n+4)\
            &= (2n^2-2n+4) - (n^2-2n+1-n+1+2)\
            &= (2n^2-2n+4) - ((n-1)^2-(n-1)+2)\
            &= (2n^2-2n+4) - frac2(n-1)^2-2(n-1)+42.
            endalign
            Do you see it now?






            share|cite|improve this answer
























              up vote
              0
              down vote













              Just rearrange to obtain
              beginalign
              n^2+n &= 2n^2-n^2 -2n+3n+4-4\
              &= (2n^2-2n+4)-(n^2-3n+4)\
              &= (2n^2-2n+4) - (n^2-2n+1-n+1+2)\
              &= (2n^2-2n+4) - ((n-1)^2-(n-1)+2)\
              &= (2n^2-2n+4) - frac2(n-1)^2-2(n-1)+42.
              endalign
              Do you see it now?






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                Just rearrange to obtain
                beginalign
                n^2+n &= 2n^2-n^2 -2n+3n+4-4\
                &= (2n^2-2n+4)-(n^2-3n+4)\
                &= (2n^2-2n+4) - (n^2-2n+1-n+1+2)\
                &= (2n^2-2n+4) - ((n-1)^2-(n-1)+2)\
                &= (2n^2-2n+4) - frac2(n-1)^2-2(n-1)+42.
                endalign
                Do you see it now?






                share|cite|improve this answer












                Just rearrange to obtain
                beginalign
                n^2+n &= 2n^2-n^2 -2n+3n+4-4\
                &= (2n^2-2n+4)-(n^2-3n+4)\
                &= (2n^2-2n+4) - (n^2-2n+1-n+1+2)\
                &= (2n^2-2n+4) - ((n-1)^2-(n-1)+2)\
                &= (2n^2-2n+4) - frac2(n-1)^2-2(n-1)+42.
                endalign
                Do you see it now?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 27 at 16:49









                Sobi

                2,418315




                2,418315



























                     

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