Finding the sum

[ref91]

I'm not sure where to start with this qs. I've calculated that U₀ =2, U₁= 1, u₂= 5/9. Where should I go from here?

 

[tkhunny]

Have you considered [math]\dfrac{2^{n}+4^{n}}{6^{n}} = \dfrac{2^{n}}{6^{n}}+\dfrac{4^{n}}{6^{n}}[/math]?

 

[Deleted member 4993]

In response #2, your problem was reduced to sum of two geometric series. What are the equations for the sum of a geometric series?

 

[ref91]

Have you considered [math]\dfrac{2^{n}+4^{n}}{6^{n}} = \dfrac{2^{n}}{6^{n}}+\dfrac{4^{n}}{6^{n}}[/math]?

Thank you

 

[ref91]

In response #2, your problem was reduced to sum of two geometric series. What are the equations for the sum of a geometric series?

thank you

 

[ref91]

Using both your help I was able to do the following:

Is that correct?

 

[tkhunny]

Please don't use finite formulations for infinite series. [math]\left(\dfrac{1}{3}\right)^{\infty}[/math] doesn't mean anything.

You got there, but it was a difficult path.

 

[Deleted member 4993]

Using both your help I was able to do the following:View attachment 22855

Is that correct?

It seems that you got a negative number for the answer (-4.5). That is incorrect. Check your formulae again!

 

[pka]

If \(|r|<1\) then \(\sum\limits_{n=K}^\infty {{r^n}} = \dfrac{{{r^K}}}{{1 - r}}\)

Thus in these questions \(\sum\limits_{n=0}^\infty {{3^{-n}}} = \dfrac{{{1}}}{{1 -3^{-1} }}=\dfrac{3}{2}\)