Find the sum of this series $sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$











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Can someone provide help in finding the sum of this series? $$sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$$






real-analysis sequences-and-series







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  • By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
    – Jack D'Aurizio
    yesterday















up vote
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down vote

favorite












Can someone provide help in finding the sum of this series? $$sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$$






real-analysis sequences-and-series







share|cite|improve this question
























  • By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
    – Jack D'Aurizio
    yesterday













up vote
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up vote
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down vote

favorite











Can someone provide help in finding the sum of this series? $$sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$$






real-analysis sequences-and-series







share|cite|improve this question















Can someone provide help in finding the sum of this series? $$sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$$






real-analysis sequences-and-series




real-analysis sequences-and-series






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edited yesterday









Lorenzo B.

1,6222419




1,6222419










asked yesterday









dimpap

657




657












  • By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
    – Jack D'Aurizio
    yesterday


















  • By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
    – Jack D'Aurizio
    yesterday
















By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
– Jack D'Aurizio
yesterday




By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
– Jack D'Aurizio
yesterday










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










Let $f(m)$



$$=dfrac{a+bm+cm^2}{2^m}$$



Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$



Use https://en.m.wikipedia.org/wiki/Telescoping_series






share|cite|improve this answer





















  • Super answer. thanks.
    – hamam_Abdallah
    yesterday










  • See also math.stackexchange.com/questions/593996/…
    – lab bhattacharjee
    yesterday










  • thanks very much!
    – dimpap
    yesterday


















up vote
0
down vote













hint



For $-2<x<2$,



$$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$



Think differentiating both sides and make $x=1$.
Observe that the numerator can be written as
$$2n(n-1)-n+4$$






share|cite|improve this answer























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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    Let $f(m)$



    $$=dfrac{a+bm+cm^2}{2^m}$$



    Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$



    Use https://en.m.wikipedia.org/wiki/Telescoping_series






    share|cite|improve this answer





















    • Super answer. thanks.
      – hamam_Abdallah
      yesterday










    • See also math.stackexchange.com/questions/593996/…
      – lab bhattacharjee
      yesterday










    • thanks very much!
      – dimpap
      yesterday















    up vote
    2
    down vote



    accepted










    Let $f(m)$



    $$=dfrac{a+bm+cm^2}{2^m}$$



    Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$



    Use https://en.m.wikipedia.org/wiki/Telescoping_series






    share|cite|improve this answer





















    • Super answer. thanks.
      – hamam_Abdallah
      yesterday










    • See also math.stackexchange.com/questions/593996/…
      – lab bhattacharjee
      yesterday










    • thanks very much!
      – dimpap
      yesterday













    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    Let $f(m)$



    $$=dfrac{a+bm+cm^2}{2^m}$$



    Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$



    Use https://en.m.wikipedia.org/wiki/Telescoping_series






    share|cite|improve this answer












    Let $f(m)$



    $$=dfrac{a+bm+cm^2}{2^m}$$



    Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$



    Use https://en.m.wikipedia.org/wiki/Telescoping_series







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    lab bhattacharjee

    220k15154270




    220k15154270












    • Super answer. thanks.
      – hamam_Abdallah
      yesterday










    • See also math.stackexchange.com/questions/593996/…
      – lab bhattacharjee
      yesterday










    • thanks very much!
      – dimpap
      yesterday


















    • Super answer. thanks.
      – hamam_Abdallah
      yesterday










    • See also math.stackexchange.com/questions/593996/…
      – lab bhattacharjee
      yesterday










    • thanks very much!
      – dimpap
      yesterday
















    Super answer. thanks.
    – hamam_Abdallah
    yesterday




    Super answer. thanks.
    – hamam_Abdallah
    yesterday












    See also math.stackexchange.com/questions/593996/…
    – lab bhattacharjee
    yesterday




    See also math.stackexchange.com/questions/593996/…
    – lab bhattacharjee
    yesterday












    thanks very much!
    – dimpap
    yesterday




    thanks very much!
    – dimpap
    yesterday










    up vote
    0
    down vote













    hint



    For $-2<x<2$,



    $$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$



    Think differentiating both sides and make $x=1$.
    Observe that the numerator can be written as
    $$2n(n-1)-n+4$$






    share|cite|improve this answer



























      up vote
      0
      down vote













      hint



      For $-2<x<2$,



      $$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$



      Think differentiating both sides and make $x=1$.
      Observe that the numerator can be written as
      $$2n(n-1)-n+4$$






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        hint



        For $-2<x<2$,



        $$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$



        Think differentiating both sides and make $x=1$.
        Observe that the numerator can be written as
        $$2n(n-1)-n+4$$






        share|cite|improve this answer














        hint



        For $-2<x<2$,



        $$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$



        Think differentiating both sides and make $x=1$.
        Observe that the numerator can be written as
        $$2n(n-1)-n+4$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        hamam_Abdallah

        36.5k21533




        36.5k21533






























             

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