The series $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+…$












1














Consider the expression $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+...$



Denote the numerator and the denominator of the $j^text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1qquad D_j=
2D_{j-1}+1$$




What is the $50^text{th}$ term?




Must we evaluate that term-by-term until we reach the $50^text{th}$ term?




What is the sum of the first $25$ term?




Must we add them one-by-one?




What is the exact value of the sum to $infty$?







sequences-and-series convergence summation recurrence-relations







share|cite|improve this question




















  • 2




    Answers to your question: no and no.
    – Yves Daoust
    Dec 9 at 9:19








  • 1




    Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
    – JimmyK4542
    Dec 9 at 9:20










  • @JimmyK4542 Yes, I did.
    – Hussain-Alqatari
    Dec 9 at 9:23










  • @AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
    – Hussain-Alqatari
    Dec 9 at 9:26










  • Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
    – Awe Kumar Jha
    Dec 9 at 9:28


















1














Consider the expression $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+...$



Denote the numerator and the denominator of the $j^text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1qquad D_j=
2D_{j-1}+1$$




What is the $50^text{th}$ term?




Must we evaluate that term-by-term until we reach the $50^text{th}$ term?




What is the sum of the first $25$ term?




Must we add them one-by-one?




What is the exact value of the sum to $infty$?







sequences-and-series convergence summation recurrence-relations







share|cite|improve this question




















  • 2




    Answers to your question: no and no.
    – Yves Daoust
    Dec 9 at 9:19








  • 1




    Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
    – JimmyK4542
    Dec 9 at 9:20










  • @JimmyK4542 Yes, I did.
    – Hussain-Alqatari
    Dec 9 at 9:23










  • @AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
    – Hussain-Alqatari
    Dec 9 at 9:26










  • Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
    – Awe Kumar Jha
    Dec 9 at 9:28
















1












1








1


1





Consider the expression $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+...$



Denote the numerator and the denominator of the $j^text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1qquad D_j=
2D_{j-1}+1$$




What is the $50^text{th}$ term?




Must we evaluate that term-by-term until we reach the $50^text{th}$ term?




What is the sum of the first $25$ term?




Must we add them one-by-one?




What is the exact value of the sum to $infty$?







sequences-and-series convergence summation recurrence-relations







share|cite|improve this question















Consider the expression $frac{1}{2}+frac{2}{5}+frac{3}{11}+frac{4}{23}+...$



Denote the numerator and the denominator of the $j^text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1qquad D_j=
2D_{j-1}+1$$




What is the $50^text{th}$ term?




Must we evaluate that term-by-term until we reach the $50^text{th}$ term?




What is the sum of the first $25$ term?




Must we add them one-by-one?




What is the exact value of the sum to $infty$?







sequences-and-series convergence summation recurrence-relations




sequences-and-series convergence summation recurrence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 9:56









Did

246k23220454




246k23220454










asked Dec 9 at 9:14









Hussain-Alqatari

2997




2997








  • 2




    Answers to your question: no and no.
    – Yves Daoust
    Dec 9 at 9:19








  • 1




    Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
    – JimmyK4542
    Dec 9 at 9:20










  • @JimmyK4542 Yes, I did.
    – Hussain-Alqatari
    Dec 9 at 9:23










  • @AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
    – Hussain-Alqatari
    Dec 9 at 9:26










  • Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
    – Awe Kumar Jha
    Dec 9 at 9:28
















  • 2




    Answers to your question: no and no.
    – Yves Daoust
    Dec 9 at 9:19








  • 1




    Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
    – JimmyK4542
    Dec 9 at 9:20










  • @JimmyK4542 Yes, I did.
    – Hussain-Alqatari
    Dec 9 at 9:23










  • @AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
    – Hussain-Alqatari
    Dec 9 at 9:26










  • Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
    – Awe Kumar Jha
    Dec 9 at 9:28










2




2




Answers to your question: no and no.
– Yves Daoust
Dec 9 at 9:19






Answers to your question: no and no.
– Yves Daoust
Dec 9 at 9:19






1




1




Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
– JimmyK4542
Dec 9 at 9:20




Did you mean to write $D_j = 2D_{j-1}+1$ for $j > 1$?
– JimmyK4542
Dec 9 at 9:20












@JimmyK4542 Yes, I did.
– Hussain-Alqatari
Dec 9 at 9:23




@JimmyK4542 Yes, I did.
– Hussain-Alqatari
Dec 9 at 9:23












@AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
– Hussain-Alqatari
Dec 9 at 9:26




@AweKumarJha , $frac{50}{523}$ is not true, the $50^text{th}$ term is much less than that.
– Hussain-Alqatari
Dec 9 at 9:26












Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
– Awe Kumar Jha
Dec 9 at 9:28






Then it just looks like a very similar series involving primes. Try using the method of generating functions to convert the recursion into explicit formula.
– Awe Kumar Jha
Dec 9 at 9:28












3 Answers
3






active

oldest

votes


















3














The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



$therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.






share|cite|improve this answer























  • This is helpful. Thank you. What about my second and the third questions?
    – Hussain-Alqatari
    Dec 9 at 9:30










  • I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
    – Anubhab Ghosal
    Dec 9 at 9:32






  • 1




    Please use cdot instead of .. The notation is really confusing.
    – Kemono Chen
    Dec 9 at 9:35










  • @KemonoChen, fixed it.
    – Anubhab Ghosal
    Dec 9 at 9:36






  • 1




    Appreciate the work, THANKS!
    – Hussain-Alqatari
    Dec 9 at 10:36



















1














If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



So 50th term is $frac{50}{3cdot2^{49}-1}$






share|cite|improve this answer





























    1














    $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



    $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




    $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



    $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



    $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



    $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



    $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



    $displaystyleimplies0<sum_1^infty s_n<2$






    share|cite|improve this answer























    • The sum is not less than $4/3$, it is $1.5997809dots$.
      – Hussain-Alqatari
      Dec 9 at 9:58










    • My bad, I'll recheck what I wrote.
      – Shubham Johri
      Dec 9 at 10:00










    • Actually, s_n > n/(3⋅2^n−1)
      – Ankit Kumar
      Dec 9 at 10:01










    • Yeah, thanks for pointing it out.
      – Shubham Johri
      Dec 9 at 10:03











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



    $therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



    I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.






    share|cite|improve this answer























    • This is helpful. Thank you. What about my second and the third questions?
      – Hussain-Alqatari
      Dec 9 at 9:30










    • I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
      – Anubhab Ghosal
      Dec 9 at 9:32






    • 1




      Please use cdot instead of .. The notation is really confusing.
      – Kemono Chen
      Dec 9 at 9:35










    • @KemonoChen, fixed it.
      – Anubhab Ghosal
      Dec 9 at 9:36






    • 1




      Appreciate the work, THANKS!
      – Hussain-Alqatari
      Dec 9 at 10:36
















    3














    The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



    $therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



    I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.






    share|cite|improve this answer























    • This is helpful. Thank you. What about my second and the third questions?
      – Hussain-Alqatari
      Dec 9 at 9:30










    • I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
      – Anubhab Ghosal
      Dec 9 at 9:32






    • 1




      Please use cdot instead of .. The notation is really confusing.
      – Kemono Chen
      Dec 9 at 9:35










    • @KemonoChen, fixed it.
      – Anubhab Ghosal
      Dec 9 at 9:36






    • 1




      Appreciate the work, THANKS!
      – Hussain-Alqatari
      Dec 9 at 10:36














    3












    3








    3






    The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



    $therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



    I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.






    share|cite|improve this answer














    The $n^{th}$ term is $frac{n}{3cdot 2^{n-1}-1}$, as can easily be proven by induction. [You can guess this by simply looking at the first two terms as you know it has to be of the form $ccdot2^n+d$]



    $therefore 50^{th}$ term$=frac{50}{3cdot 2^{49}-1}$.



    I tried all the methods I know(which includes generating functions, bruteforce calculation, CAS). The sum does not have a closed form formula.(As is the case with most nontrivial rapidly converging sums). You may however calculate the sum with arbitrary precision pretty easily.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 at 10:22

























    answered Dec 9 at 9:28









    Anubhab Ghosal

    74412




    74412












    • This is helpful. Thank you. What about my second and the third questions?
      – Hussain-Alqatari
      Dec 9 at 9:30










    • I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
      – Anubhab Ghosal
      Dec 9 at 9:32






    • 1




      Please use cdot instead of .. The notation is really confusing.
      – Kemono Chen
      Dec 9 at 9:35










    • @KemonoChen, fixed it.
      – Anubhab Ghosal
      Dec 9 at 9:36






    • 1




      Appreciate the work, THANKS!
      – Hussain-Alqatari
      Dec 9 at 10:36


















    • This is helpful. Thank you. What about my second and the third questions?
      – Hussain-Alqatari
      Dec 9 at 9:30










    • I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
      – Anubhab Ghosal
      Dec 9 at 9:32






    • 1




      Please use cdot instead of .. The notation is really confusing.
      – Kemono Chen
      Dec 9 at 9:35










    • @KemonoChen, fixed it.
      – Anubhab Ghosal
      Dec 9 at 9:36






    • 1




      Appreciate the work, THANKS!
      – Hussain-Alqatari
      Dec 9 at 10:36
















    This is helpful. Thank you. What about my second and the third questions?
    – Hussain-Alqatari
    Dec 9 at 9:30




    This is helpful. Thank you. What about my second and the third questions?
    – Hussain-Alqatari
    Dec 9 at 9:30












    I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
    – Anubhab Ghosal
    Dec 9 at 9:32




    I am trying those parts. Will edit as soon as I get something useful. As of now, I think there is no closed form for the summation.
    – Anubhab Ghosal
    Dec 9 at 9:32




    1




    1




    Please use cdot instead of .. The notation is really confusing.
    – Kemono Chen
    Dec 9 at 9:35




    Please use cdot instead of .. The notation is really confusing.
    – Kemono Chen
    Dec 9 at 9:35












    @KemonoChen, fixed it.
    – Anubhab Ghosal
    Dec 9 at 9:36




    @KemonoChen, fixed it.
    – Anubhab Ghosal
    Dec 9 at 9:36




    1




    1




    Appreciate the work, THANKS!
    – Hussain-Alqatari
    Dec 9 at 10:36




    Appreciate the work, THANKS!
    – Hussain-Alqatari
    Dec 9 at 10:36











    1














    If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



    So 50th term is $frac{50}{3cdot2^{49}-1}$






    share|cite|improve this answer


























      1














      If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



      So 50th term is $frac{50}{3cdot2^{49}-1}$






      share|cite|improve this answer
























        1












        1








        1






        If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



        So 50th term is $frac{50}{3cdot2^{49}-1}$






        share|cite|improve this answer












        If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)Rightarrow d_n+1=(d_1+1)2^{n-1}=3cdot2^{n-1}$.



        So 50th term is $frac{50}{3cdot2^{49}-1}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 at 9:39









        Minz

        951127




        951127























            1














            $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



            $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




            $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



            $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



            $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



            $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



            $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



            $displaystyleimplies0<sum_1^infty s_n<2$






            share|cite|improve this answer























            • The sum is not less than $4/3$, it is $1.5997809dots$.
              – Hussain-Alqatari
              Dec 9 at 9:58










            • My bad, I'll recheck what I wrote.
              – Shubham Johri
              Dec 9 at 10:00










            • Actually, s_n > n/(3⋅2^n−1)
              – Ankit Kumar
              Dec 9 at 10:01










            • Yeah, thanks for pointing it out.
              – Shubham Johri
              Dec 9 at 10:03
















            1














            $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



            $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




            $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



            $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



            $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



            $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



            $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



            $displaystyleimplies0<sum_1^infty s_n<2$






            share|cite|improve this answer























            • The sum is not less than $4/3$, it is $1.5997809dots$.
              – Hussain-Alqatari
              Dec 9 at 9:58










            • My bad, I'll recheck what I wrote.
              – Shubham Johri
              Dec 9 at 10:00










            • Actually, s_n > n/(3⋅2^n−1)
              – Ankit Kumar
              Dec 9 at 10:01










            • Yeah, thanks for pointing it out.
              – Shubham Johri
              Dec 9 at 10:03














            1












            1








            1






            $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



            $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




            $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



            $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



            $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



            $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



            $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



            $displaystyleimplies0<sum_1^infty s_n<2$






            share|cite|improve this answer














            $D_j=2D_{j-1}+1\ =2(2D_{j-2}+1)+1\ =4D_{j-2}+1+2\ =4(2D_{j-3}+1)+1+2\ vdots\ =2^kD_{j-k}+2^k-1\ =2^{j-1}D_1+2^{j-1}-1\ =3cdot2^{j-1}-1$



            $s_n=N_n/D_n=displaystylefrac n{3cdot2^{n-1}-1}, nge1$




            $s_n<displaystylefrac n{3cdot2^{n-1}-2^{n-1}}=frac n{2^n}$



            $displaystylesum_1^infty s_n<sum_1^infty frac n{2^n}$ which is an AP-GP series



            $sum_1^infty frac n{2^n}=frac12+frac24+frac38...$



            $frac12sum_1^infty frac n{2^n}=0+frac14+frac28+frac3{16}...$



            $sum_1^infty frac n{2^n}-frac12sum_1^infty frac n{2^n}=frac12sum_1^infty frac n{2^n}=frac12+frac14+frac18...=1$



            $displaystyleimplies0<sum_1^infty s_n<2$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 9 at 10:18

























            answered Dec 9 at 9:52









            Shubham Johri

            3,826716




            3,826716












            • The sum is not less than $4/3$, it is $1.5997809dots$.
              – Hussain-Alqatari
              Dec 9 at 9:58










            • My bad, I'll recheck what I wrote.
              – Shubham Johri
              Dec 9 at 10:00










            • Actually, s_n > n/(3⋅2^n−1)
              – Ankit Kumar
              Dec 9 at 10:01










            • Yeah, thanks for pointing it out.
              – Shubham Johri
              Dec 9 at 10:03


















            • The sum is not less than $4/3$, it is $1.5997809dots$.
              – Hussain-Alqatari
              Dec 9 at 9:58










            • My bad, I'll recheck what I wrote.
              – Shubham Johri
              Dec 9 at 10:00










            • Actually, s_n > n/(3⋅2^n−1)
              – Ankit Kumar
              Dec 9 at 10:01










            • Yeah, thanks for pointing it out.
              – Shubham Johri
              Dec 9 at 10:03
















            The sum is not less than $4/3$, it is $1.5997809dots$.
            – Hussain-Alqatari
            Dec 9 at 9:58




            The sum is not less than $4/3$, it is $1.5997809dots$.
            – Hussain-Alqatari
            Dec 9 at 9:58












            My bad, I'll recheck what I wrote.
            – Shubham Johri
            Dec 9 at 10:00




            My bad, I'll recheck what I wrote.
            – Shubham Johri
            Dec 9 at 10:00












            Actually, s_n > n/(3⋅2^n−1)
            – Ankit Kumar
            Dec 9 at 10:01




            Actually, s_n > n/(3⋅2^n−1)
            – Ankit Kumar
            Dec 9 at 10:01












            Yeah, thanks for pointing it out.
            – Shubham Johri
            Dec 9 at 10:03




            Yeah, thanks for pointing it out.
            – Shubham Johri
            Dec 9 at 10:03


















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