Respuesta :

LammettHash
If you're talking about adding just these four terms, notice that

[tex]S_4=6+2+\dfrac23+\dfrac29[/tex]
[tex]S_4=6\left(1+\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}\right)[/tex]
[tex]\dfrac13S_4=6\left(\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}\right)[/tex]

[tex]\implies S_4-\dfrac13S_4=6\left(1-\dfrac1{3^4}\right)
[tex]\dfrac23S_4=6\left(1-\dfrac1{81}\right)[/tex]
[tex]S_4=9\times\dfrac{80}{81}[/tex]
[tex]S_4=\dfrac{80}9[/tex]

If you were referring to an infinite sum, so that the pattern continues, note that the same procedure as above can be applied to the [tex]n[/tex]th partial sum:

[tex]S_n=6\left(1+\dfrac13+\dfrac1{3^2}+\cdots+\dfrac1{3^n}\right)[/tex]
[tex]\dfrac23S_n=6\left(1-\dfrac1{3^{n+1}}\right)[/tex]
[tex]S_n=9\left(1-\dfrac1{3^{n+1}}\right)[/tex]

Then as [tex]n\to\infty[/tex], the exponential term approaches 0, leaving you with

[tex]6+2+\dfrac23+\dfrac29+\cdots=9[/tex]

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