Sum of fraction series












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I have been trying to solve this question but I am not sure how to, and would appreciate some help. Thanks.




Simplify:
$$frac{1}{4} +frac{1}{4^2} +frac{4^2}{4^3} +frac{4^2}{4^4} +frac{4^4}{4^5} +frac{4^4}{4^6} +ldots +frac{4^{102}}{4^{103}} +frac{4^{102}}{4^{104}}.$$







fractions







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  • $begingroup$
    Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
    $endgroup$
    – J.G.
    Dec 8 '18 at 14:21






  • 4




    $begingroup$
    You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
    $endgroup$
    – KM101
    Dec 8 '18 at 14:23


















0












$begingroup$


I have been trying to solve this question but I am not sure how to, and would appreciate some help. Thanks.




Simplify:
$$frac{1}{4} +frac{1}{4^2} +frac{4^2}{4^3} +frac{4^2}{4^4} +frac{4^4}{4^5} +frac{4^4}{4^6} +ldots +frac{4^{102}}{4^{103}} +frac{4^{102}}{4^{104}}.$$







fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
    $endgroup$
    – J.G.
    Dec 8 '18 at 14:21






  • 4




    $begingroup$
    You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
    $endgroup$
    – KM101
    Dec 8 '18 at 14:23
















0












0








0





$begingroup$


I have been trying to solve this question but I am not sure how to, and would appreciate some help. Thanks.




Simplify:
$$frac{1}{4} +frac{1}{4^2} +frac{4^2}{4^3} +frac{4^2}{4^4} +frac{4^4}{4^5} +frac{4^4}{4^6} +ldots +frac{4^{102}}{4^{103}} +frac{4^{102}}{4^{104}}.$$







fractions







share|cite|improve this question











$endgroup$




I have been trying to solve this question but I am not sure how to, and would appreciate some help. Thanks.




Simplify:
$$frac{1}{4} +frac{1}{4^2} +frac{4^2}{4^3} +frac{4^2}{4^4} +frac{4^4}{4^5} +frac{4^4}{4^6} +ldots +frac{4^{102}}{4^{103}} +frac{4^{102}}{4^{104}}.$$







fractions




fractions






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share|cite|improve this question













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share|cite|improve this question








edited Dec 8 '18 at 14:19









Rócherz

2,7762721




2,7762721










asked Dec 8 '18 at 14:16









UnknownUnknown

73




73












  • $begingroup$
    Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
    $endgroup$
    – J.G.
    Dec 8 '18 at 14:21






  • 4




    $begingroup$
    You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
    $endgroup$
    – KM101
    Dec 8 '18 at 14:23




















  • $begingroup$
    Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
    $endgroup$
    – J.G.
    Dec 8 '18 at 14:21






  • 4




    $begingroup$
    You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
    $endgroup$
    – KM101
    Dec 8 '18 at 14:23


















$begingroup$
Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
$endgroup$
– J.G.
Dec 8 '18 at 14:21




$begingroup$
Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
$endgroup$
– J.G.
Dec 8 '18 at 14:21




4




4




$begingroup$
You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
$endgroup$
– KM101
Dec 8 '18 at 14:23






$begingroup$
You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
$endgroup$
– KM101
Dec 8 '18 at 14:23












1 Answer
1






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$begingroup$

The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
$$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
$$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
$$=13+frac{13}{4}$$
$$=frac{65}{4}$$
Hope it is helpful:)






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






    active

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    oldest

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    active

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    0












    $begingroup$

    The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
    $$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
    $$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
    $$=13+frac{13}{4}$$
    $$=frac{65}{4}$$
    Hope it is helpful:)






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
      $$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
      $$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
      $$=13+frac{13}{4}$$
      $$=frac{65}{4}$$
      Hope it is helpful:)






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
        $$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
        $$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
        $$=13+frac{13}{4}$$
        $$=frac{65}{4}$$
        Hope it is helpful:)






        share|cite|improve this answer









        $endgroup$



        The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
        $$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
        $$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
        $$=13+frac{13}{4}$$
        $$=frac{65}{4}$$
        Hope it is helpful:)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 14:53









        MartundMartund

        1,593212




        1,593212






























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