Double summation with a variable stopping point [closed]

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I am interested in calculating the following double summation:



$sum_n=2^ infty sum_k =0^n-2frac14^k frac12^n-k-2$



I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.



The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.




sequences-and-series summation terminology




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closed as off-topic by amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco Jul 31 at 8:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn♦
    Jul 31 at 0:22










  • The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context. I will make sure to do so in the future.
    – Filip
    Jul 31 at 2:13










  • I have added the comment as context to your question. Thank you for doing this in the future.
    – robjohn♦
    Jul 31 at 3:14














up vote
1
down vote

favorite












I am interested in calculating the following double summation:



$sum_n=2^ infty sum_k =0^n-2frac14^k frac12^n-k-2$



I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.



The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.




sequences-and-series summation terminology




share|cite|improve this question













closed as off-topic by amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco Jul 31 at 8:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn♦
    Jul 31 at 0:22










  • The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context. I will make sure to do so in the future.
    – Filip
    Jul 31 at 2:13










  • I have added the comment as context to your question. Thank you for doing this in the future.
    – robjohn♦
    Jul 31 at 3:14












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I am interested in calculating the following double summation:



$sum_n=2^ infty sum_k =0^n-2frac14^k frac12^n-k-2$



I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.



The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.




sequences-and-series summation terminology




share|cite|improve this question













I am interested in calculating the following double summation:



$sum_n=2^ infty sum_k =0^n-2frac14^k frac12^n-k-2$



I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.



The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.





sequences-and-series summation terminology





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 3:13









robjohn♦

258k25296612




258k25296612









asked Jul 30 at 23:54









Filip

15119




15119




closed as off-topic by amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco Jul 31 at 8:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco Jul 31 at 8:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn♦
    Jul 31 at 0:22










  • The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context. I will make sure to do so in the future.
    – Filip
    Jul 31 at 2:13










  • I have added the comment as context to your question. Thank you for doing this in the future.
    – robjohn♦
    Jul 31 at 3:14
















  • Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn♦
    Jul 31 at 0:22










  • The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context. I will make sure to do so in the future.
    – Filip
    Jul 31 at 2:13










  • I have added the comment as context to your question. Thank you for doing this in the future.
    – robjohn♦
    Jul 31 at 3:14















Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
– robjohn♦
Jul 31 at 0:22




Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
– robjohn♦
Jul 31 at 0:22












The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context. I will make sure to do so in the future.
– Filip
Jul 31 at 2:13




The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context. I will make sure to do so in the future.
– Filip
Jul 31 at 2:13












I have added the comment as context to your question. Thank you for doing this in the future.
– robjohn♦
Jul 31 at 3:14




I have added the comment as context to your question. Thank you for doing this in the future.
– robjohn♦
Jul 31 at 3:14










3 Answers
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$$
beginalign
sum_n=2^inftysum_k=0^n-2left(frac14right)^kleft(frac12right)^n-k-2
&=sum_n=0^inftysum_k=0^nleft(frac14right)^kleft(frac12right)^n-ktag1\
&=sum_k=0^inftysum_n=k^inftyleft(frac14right)^kleft(frac12right)^n-ktag2\
&=sum_k=0^inftysum_n=0^inftyleft(frac14right)^kleft(frac12right)^ntag3\
endalign
$$
Explanation:

$(1)$: substitute $nmapsto n+2$

$(2)$: switch order of summation ($nge k$)

$(3)$: substitute $nmapsto n+k$



Does this look easier?






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    beginaligned
    sum_n=2^ infty sum_k =0^n-2left(frac14right)^k left(frac12right)^n-k-2 &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^2k left(frac12right)^n-k-2 \
    &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^n+k-2 \
    &= sum_n=2^ infty left(frac12right)^n-2sum_k =0^n-2left(frac12right)^k \
    &= sum_n=2^ infty left(frac12right)^n-2 frac1-left(frac 12right)^n-11-frac 12 \
    &= sum_n=2^ inftyleft(frac12right)^n-3 - left(frac12right)^2n-4 \
    &= 2 sum_n=0^ inftyleft(frac12right)^n -sum_n=0^inftyleft(frac14right)^n \
    &= 2 frac11-frac 12-frac11-frac 1 4 \
    &= frac 8 3
    endaligned






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      This problem involves sums of geometric sequences for which there is a well-known formula:



      $$sum_n=0^m-1 r^n = frac1-r^m1-r.$$



      Applying this to your summation and using a little algebra you get:



      $$beginequation beginaligned
      sum_n=2^infty sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^n-k-2
      &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^-k \[6pt]
      &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac12 Big)^k \[6pt]
      &= sum_n=2^infty Big( frac12 Big)^n-2 cdot frac1-(tfrac12)^n-11-tfrac12 \[6pt]
      &= 2 Bigg[ sum_n=2^infty Big( frac12 Big)^n-2 - sum_n=2^infty Big( frac12 Big)^2n-3 Bigg] \[6pt]
      &= 2 Bigg[ sum_n=0^infty Big( frac12 Big)^n - frac12 sum_n=0^infty Big( frac14 Big)^n Bigg] \[6pt]
      &= 2 Bigg[ frac11-tfrac12 - frac12 cdot frac11-tfrac14 Bigg] \[6pt]
      &= 2 Bigg[ 2 - frac12 cdot frac43 Bigg] \[6pt]
      &= 2 Bigg[ 2 - frac23 Bigg] \[6pt]
      &= 2 cdot frac43 \[6pt]
      &= frac83 = 2.6bar6. \[6pt]
      endaligned endequation$$






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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
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        down vote













        $$
        beginalign
        sum_n=2^inftysum_k=0^n-2left(frac14right)^kleft(frac12right)^n-k-2
        &=sum_n=0^inftysum_k=0^nleft(frac14right)^kleft(frac12right)^n-ktag1\
        &=sum_k=0^inftysum_n=k^inftyleft(frac14right)^kleft(frac12right)^n-ktag2\
        &=sum_k=0^inftysum_n=0^inftyleft(frac14right)^kleft(frac12right)^ntag3\
        endalign
        $$
        Explanation:

        $(1)$: substitute $nmapsto n+2$

        $(2)$: switch order of summation ($nge k$)

        $(3)$: substitute $nmapsto n+k$



        Does this look easier?






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          up vote
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          down vote













          $$
          beginalign
          sum_n=2^inftysum_k=0^n-2left(frac14right)^kleft(frac12right)^n-k-2
          &=sum_n=0^inftysum_k=0^nleft(frac14right)^kleft(frac12right)^n-ktag1\
          &=sum_k=0^inftysum_n=k^inftyleft(frac14right)^kleft(frac12right)^n-ktag2\
          &=sum_k=0^inftysum_n=0^inftyleft(frac14right)^kleft(frac12right)^ntag3\
          endalign
          $$
          Explanation:

          $(1)$: substitute $nmapsto n+2$

          $(2)$: switch order of summation ($nge k$)

          $(3)$: substitute $nmapsto n+k$



          Does this look easier?






          share|cite|improve this answer























            up vote
            0
            down vote










            up vote
            0
            down vote









            $$
            beginalign
            sum_n=2^inftysum_k=0^n-2left(frac14right)^kleft(frac12right)^n-k-2
            &=sum_n=0^inftysum_k=0^nleft(frac14right)^kleft(frac12right)^n-ktag1\
            &=sum_k=0^inftysum_n=k^inftyleft(frac14right)^kleft(frac12right)^n-ktag2\
            &=sum_k=0^inftysum_n=0^inftyleft(frac14right)^kleft(frac12right)^ntag3\
            endalign
            $$
            Explanation:

            $(1)$: substitute $nmapsto n+2$

            $(2)$: switch order of summation ($nge k$)

            $(3)$: substitute $nmapsto n+k$



            Does this look easier?






            share|cite|improve this answer













            $$
            beginalign
            sum_n=2^inftysum_k=0^n-2left(frac14right)^kleft(frac12right)^n-k-2
            &=sum_n=0^inftysum_k=0^nleft(frac14right)^kleft(frac12right)^n-ktag1\
            &=sum_k=0^inftysum_n=k^inftyleft(frac14right)^kleft(frac12right)^n-ktag2\
            &=sum_k=0^inftysum_n=0^inftyleft(frac14right)^kleft(frac12right)^ntag3\
            endalign
            $$
            Explanation:

            $(1)$: substitute $nmapsto n+2$

            $(2)$: switch order of summation ($nge k$)

            $(3)$: substitute $nmapsto n+k$



            Does this look easier?







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 31 at 0:04









            robjohn♦

            258k25296612




            258k25296612




















                up vote
                0
                down vote













                beginaligned
                sum_n=2^ infty sum_k =0^n-2left(frac14right)^k left(frac12right)^n-k-2 &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^2k left(frac12right)^n-k-2 \
                &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^n+k-2 \
                &= sum_n=2^ infty left(frac12right)^n-2sum_k =0^n-2left(frac12right)^k \
                &= sum_n=2^ infty left(frac12right)^n-2 frac1-left(frac 12right)^n-11-frac 12 \
                &= sum_n=2^ inftyleft(frac12right)^n-3 - left(frac12right)^2n-4 \
                &= 2 sum_n=0^ inftyleft(frac12right)^n -sum_n=0^inftyleft(frac14right)^n \
                &= 2 frac11-frac 12-frac11-frac 1 4 \
                &= frac 8 3
                endaligned






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                  beginaligned
                  sum_n=2^ infty sum_k =0^n-2left(frac14right)^k left(frac12right)^n-k-2 &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^2k left(frac12right)^n-k-2 \
                  &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^n+k-2 \
                  &= sum_n=2^ infty left(frac12right)^n-2sum_k =0^n-2left(frac12right)^k \
                  &= sum_n=2^ infty left(frac12right)^n-2 frac1-left(frac 12right)^n-11-frac 12 \
                  &= sum_n=2^ inftyleft(frac12right)^n-3 - left(frac12right)^2n-4 \
                  &= 2 sum_n=0^ inftyleft(frac12right)^n -sum_n=0^inftyleft(frac14right)^n \
                  &= 2 frac11-frac 12-frac11-frac 1 4 \
                  &= frac 8 3
                  endaligned






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                    up vote
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                    down vote










                    up vote
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                    down vote









                    beginaligned
                    sum_n=2^ infty sum_k =0^n-2left(frac14right)^k left(frac12right)^n-k-2 &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^2k left(frac12right)^n-k-2 \
                    &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^n+k-2 \
                    &= sum_n=2^ infty left(frac12right)^n-2sum_k =0^n-2left(frac12right)^k \
                    &= sum_n=2^ infty left(frac12right)^n-2 frac1-left(frac 12right)^n-11-frac 12 \
                    &= sum_n=2^ inftyleft(frac12right)^n-3 - left(frac12right)^2n-4 \
                    &= 2 sum_n=0^ inftyleft(frac12right)^n -sum_n=0^inftyleft(frac14right)^n \
                    &= 2 frac11-frac 12-frac11-frac 1 4 \
                    &= frac 8 3
                    endaligned






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                    beginaligned
                    sum_n=2^ infty sum_k =0^n-2left(frac14right)^k left(frac12right)^n-k-2 &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^2k left(frac12right)^n-k-2 \
                    &= sum_n=2^ infty sum_k =0^n-2left(frac12right)^n+k-2 \
                    &= sum_n=2^ infty left(frac12right)^n-2sum_k =0^n-2left(frac12right)^k \
                    &= sum_n=2^ infty left(frac12right)^n-2 frac1-left(frac 12right)^n-11-frac 12 \
                    &= sum_n=2^ inftyleft(frac12right)^n-3 - left(frac12right)^2n-4 \
                    &= 2 sum_n=0^ inftyleft(frac12right)^n -sum_n=0^inftyleft(frac14right)^n \
                    &= 2 frac11-frac 12-frac11-frac 1 4 \
                    &= frac 8 3
                    endaligned







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                    answered Jul 31 at 0:24









                    Ali

                    1,144924




                    1,144924




















                        up vote
                        0
                        down vote













                        This problem involves sums of geometric sequences for which there is a well-known formula:



                        $$sum_n=0^m-1 r^n = frac1-r^m1-r.$$



                        Applying this to your summation and using a little algebra you get:



                        $$beginequation beginaligned
                        sum_n=2^infty sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^n-k-2
                        &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^-k \[6pt]
                        &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac12 Big)^k \[6pt]
                        &= sum_n=2^infty Big( frac12 Big)^n-2 cdot frac1-(tfrac12)^n-11-tfrac12 \[6pt]
                        &= 2 Bigg[ sum_n=2^infty Big( frac12 Big)^n-2 - sum_n=2^infty Big( frac12 Big)^2n-3 Bigg] \[6pt]
                        &= 2 Bigg[ sum_n=0^infty Big( frac12 Big)^n - frac12 sum_n=0^infty Big( frac14 Big)^n Bigg] \[6pt]
                        &= 2 Bigg[ frac11-tfrac12 - frac12 cdot frac11-tfrac14 Bigg] \[6pt]
                        &= 2 Bigg[ 2 - frac12 cdot frac43 Bigg] \[6pt]
                        &= 2 Bigg[ 2 - frac23 Bigg] \[6pt]
                        &= 2 cdot frac43 \[6pt]
                        &= frac83 = 2.6bar6. \[6pt]
                        endaligned endequation$$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          This problem involves sums of geometric sequences for which there is a well-known formula:



                          $$sum_n=0^m-1 r^n = frac1-r^m1-r.$$



                          Applying this to your summation and using a little algebra you get:



                          $$beginequation beginaligned
                          sum_n=2^infty sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^n-k-2
                          &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^-k \[6pt]
                          &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac12 Big)^k \[6pt]
                          &= sum_n=2^infty Big( frac12 Big)^n-2 cdot frac1-(tfrac12)^n-11-tfrac12 \[6pt]
                          &= 2 Bigg[ sum_n=2^infty Big( frac12 Big)^n-2 - sum_n=2^infty Big( frac12 Big)^2n-3 Bigg] \[6pt]
                          &= 2 Bigg[ sum_n=0^infty Big( frac12 Big)^n - frac12 sum_n=0^infty Big( frac14 Big)^n Bigg] \[6pt]
                          &= 2 Bigg[ frac11-tfrac12 - frac12 cdot frac11-tfrac14 Bigg] \[6pt]
                          &= 2 Bigg[ 2 - frac12 cdot frac43 Bigg] \[6pt]
                          &= 2 Bigg[ 2 - frac23 Bigg] \[6pt]
                          &= 2 cdot frac43 \[6pt]
                          &= frac83 = 2.6bar6. \[6pt]
                          endaligned endequation$$






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            This problem involves sums of geometric sequences for which there is a well-known formula:



                            $$sum_n=0^m-1 r^n = frac1-r^m1-r.$$



                            Applying this to your summation and using a little algebra you get:



                            $$beginequation beginaligned
                            sum_n=2^infty sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^n-k-2
                            &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^-k \[6pt]
                            &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac12 Big)^k \[6pt]
                            &= sum_n=2^infty Big( frac12 Big)^n-2 cdot frac1-(tfrac12)^n-11-tfrac12 \[6pt]
                            &= 2 Bigg[ sum_n=2^infty Big( frac12 Big)^n-2 - sum_n=2^infty Big( frac12 Big)^2n-3 Bigg] \[6pt]
                            &= 2 Bigg[ sum_n=0^infty Big( frac12 Big)^n - frac12 sum_n=0^infty Big( frac14 Big)^n Bigg] \[6pt]
                            &= 2 Bigg[ frac11-tfrac12 - frac12 cdot frac11-tfrac14 Bigg] \[6pt]
                            &= 2 Bigg[ 2 - frac12 cdot frac43 Bigg] \[6pt]
                            &= 2 Bigg[ 2 - frac23 Bigg] \[6pt]
                            &= 2 cdot frac43 \[6pt]
                            &= frac83 = 2.6bar6. \[6pt]
                            endaligned endequation$$






                            share|cite|improve this answer













                            This problem involves sums of geometric sequences for which there is a well-known formula:



                            $$sum_n=0^m-1 r^n = frac1-r^m1-r.$$



                            Applying this to your summation and using a little algebra you get:



                            $$beginequation beginaligned
                            sum_n=2^infty sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^n-k-2
                            &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac14 Big)^k Big( frac12 Big)^-k \[6pt]
                            &= sum_n=2^infty Big( frac12 Big)^n-2 sum_k=0^n-2 Big( frac12 Big)^k \[6pt]
                            &= sum_n=2^infty Big( frac12 Big)^n-2 cdot frac1-(tfrac12)^n-11-tfrac12 \[6pt]
                            &= 2 Bigg[ sum_n=2^infty Big( frac12 Big)^n-2 - sum_n=2^infty Big( frac12 Big)^2n-3 Bigg] \[6pt]
                            &= 2 Bigg[ sum_n=0^infty Big( frac12 Big)^n - frac12 sum_n=0^infty Big( frac14 Big)^n Bigg] \[6pt]
                            &= 2 Bigg[ frac11-tfrac12 - frac12 cdot frac11-tfrac14 Bigg] \[6pt]
                            &= 2 Bigg[ 2 - frac12 cdot frac43 Bigg] \[6pt]
                            &= 2 Bigg[ 2 - frac23 Bigg] \[6pt]
                            &= 2 cdot frac43 \[6pt]
                            &= frac83 = 2.6bar6. \[6pt]
                            endaligned endequation$$







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                            answered Jul 31 at 0:29









                            Ben

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