Respuesta :

Answer:

[tex]\large \boxed{\ \ \dfrac{63}{5} \ \ }[/tex]

Step-by-step explanation:

Hello,

"Find the sum of the following infinite geometric series"

infinite

   We will have to find the limit of something when n tends to [tex]+\infty[/tex]

geometric series

   This is a good clue, meaning that each term of the series follows a geometric sequence. Let's check that.

The sum is something like

           [tex]\displaystyle \sum_{k=0}^{+\infty} a_k[/tex]

First of all, we need to find an expression for [tex]a_k[/tex]

First term is

   [tex]a_0=7[/tex]

Second term is

    [tex]a_1=\dfrac{4}{9}\cdot a_0=7*\boxed{\dfrac{4}{9}}=\dfrac{7*4}{9}=\dfrac{28}{9}[/tex]

Then

   [tex]a_2=\dfrac{4}{9}\cdot a_1=\dfrac{28}{9}*\boxed{\dfrac{4}{9}}=\dfrac{28*4}{9*9}=\dfrac{112}{81}[/tex]

and...

   [tex]a_3=\dfrac{4}{9}\cdot a_2=\dfrac{112}{81}*\boxed{\dfrac{4}{9}}=\dfrac{112*4}{9*81}=\dfrac{448}{729}[/tex]

Ok we are good, we can express any term for k integer

   [tex]a_k=a_0\cdot (\dfrac{4}{9})^k[/tex]

So, for n positive integer

[tex]\displaystyle \sum_{k=0}^{n} a_k=\displaystyle \sum_{k=0}^{n} 7\cdot (\dfrac{4}{9})^k=7\cdot \dfrac{1-(\dfrac{4}{9})^{n+1}}{1-\dfrac{4}{9}}=\dfrac{7*9*[1-(\dfrac{4}{9})^{n+1}]}{9-4}=\dfrac{63}{5}\cdot [1-(\dfrac{4}{9})^{n+1}}][/tex]

And the limit of that expression when n tends to [tex]+\infty[/tex] is

   [tex]\large \boxed{\ \ \dfrac{63}{5} \ \ }[/tex]

as

   [tex]\dfrac{4}{9}<1[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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