How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $












2












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How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ sum_{n=1}^{infty}(2^n cdot (frac{2}{11})^n+2cdot(frac{2}{11})^ncdot(-1)^n + frac{1}{11^n})$$ But what should be done after...?






real-analysis







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  • $begingroup$
    Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
    $endgroup$
    – Jean-Claude Arbaut
    Nov 25 '18 at 21:38












  • $begingroup$
    Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
    $endgroup$
    – VirtualUser
    Nov 25 '18 at 21:41


















2












$begingroup$


How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ sum_{n=1}^{infty}(2^n cdot (frac{2}{11})^n+2cdot(frac{2}{11})^ncdot(-1)^n + frac{1}{11^n})$$ But what should be done after...?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
    $endgroup$
    – Jean-Claude Arbaut
    Nov 25 '18 at 21:38












  • $begingroup$
    Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
    $endgroup$
    – VirtualUser
    Nov 25 '18 at 21:41
















2












2








2





$begingroup$


How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ sum_{n=1}^{infty}(2^n cdot (frac{2}{11})^n+2cdot(frac{2}{11})^ncdot(-1)^n + frac{1}{11^n})$$ But what should be done after...?






real-analysis







share|cite|improve this question











$endgroup$




How to calculate $sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ sum_{n=1}^{infty}(2^n cdot (frac{2}{11})^n+2cdot(frac{2}{11})^ncdot(-1)^n + frac{1}{11^n})$$ But what should be done after...?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 21:34







VirtualUser

















asked Nov 25 '18 at 21:22









VirtualUserVirtualUser

63912




63912












  • $begingroup$
    Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
    $endgroup$
    – Jean-Claude Arbaut
    Nov 25 '18 at 21:38












  • $begingroup$
    Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
    $endgroup$
    – VirtualUser
    Nov 25 '18 at 21:41




















  • $begingroup$
    Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
    $endgroup$
    – Jean-Claude Arbaut
    Nov 25 '18 at 21:38












  • $begingroup$
    Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
    $endgroup$
    – VirtualUser
    Nov 25 '18 at 21:41


















$begingroup$
Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
$endgroup$
– Jean-Claude Arbaut
Nov 25 '18 at 21:38






$begingroup$
Now you have $3$ geometric series, all convergent, and you can use the fact that for $|p|<1$, $$sum_{p=0}^infty p^n=frac{1}{1-p}$$
$endgroup$
– Jean-Claude Arbaut
Nov 25 '18 at 21:38














$begingroup$
Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
$endgroup$
– VirtualUser
Nov 25 '18 at 21:41






$begingroup$
Best answer! Repeat that as normal answer and I will mark that as the best :) @Jean-ClaudeArbaut
$endgroup$
– VirtualUser
Nov 25 '18 at 21:41












2 Answers
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$begingroup$

For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



$$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
= frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$






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    0












    $begingroup$

    Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



    $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



    These are just geometric series, so you can apply the formula



    $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
      2






      active

      oldest

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      active

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      active

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      2












      $begingroup$

      For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



      $$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
      = frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



        $$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
        = frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



          $$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
          = frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$






          share|cite|improve this answer









          $endgroup$



          For $|p|<1$, $sum_{n=1}^{infty}p^n=frac{p}{1-p}$, so



          $$sum_{n=1}^{infty}frac{(2^n+(-1)^n)^2}{11^n}=sum_{n=1}^{infty}frac{4^n}{11^n}+2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}\
          = frac{frac{4}{11}}{1-frac{4}{11}}-2frac{frac{2}{11}}{1+frac{2}{11}}+frac{frac{1}{11}}{1-frac{1}{11}}\=frac{4}{7}-frac{4}{13}+frac{1}{10}=frac{331}{910}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 '18 at 21:44









          Jean-Claude ArbautJean-Claude Arbaut

          14.7k63464




          14.7k63464























              0












              $begingroup$

              Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



              $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



              These are just geometric series, so you can apply the formula



              $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



                $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



                These are just geometric series, so you can apply the formula



                $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



                  $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



                  These are just geometric series, so you can apply the formula



                  $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$






                  share|cite|improve this answer









                  $endgroup$



                  Once you've expanded $(2^n-(-1)^n)^2=4^n-2(-2)^n+1$, split up the summation as



                  $$sum_{n=1}^{infty} frac{4^n}{11^n}-2sum_{n=1}^{infty}frac{(-2)^n}{11^n}+sum_{n=1}^{infty}frac{1}{11^n}.$$



                  These are just geometric series, so you can apply the formula



                  $$sum_{n=1}^{infty} x^n=frac{x}{1-x}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 '18 at 21:45









                  Carl SchildkrautCarl Schildkraut

                  11.2k11441




                  11.2k11441






























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