what is the sum of the following series:
infinity E n=1 ((2^n+8^n)/11^n))

Question

what is the sum of the following series:
 infinity E n=1 ((2^n+8^n)/11^n))

solomonzelman · Answer

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 2^n+8^n }{11^n}}$

like this?

anonymous · Answer

yes
Im really lost ... :(

solomonzelman · Answer

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 2^n+8^n }{11^n}}$

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 2^n }{11^n}+\frac{ 8^n }{11^n}}$

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 2^n }{11^n}+\sum_{ n=1 }^{ \infty } \frac{ 8^n }{11^n}}$

solomonzelman · Answer

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \left(\frac{ 2 }{11}\right)^n+\sum_{ n=1 }^{ \infty } ~ \left(\frac{ 8 }{11}\right)^n}$

solomonzelman · Answer

Do you know about "geometric series" ?

anonymous · Answer

I think i understand it a bit. its =A/1-x?

solomonzelman · Answer

$a_1~/~(1-r)$

where r is the common ratio
and $a_1$ is the first term of the series.

solomonzelman · Answer

Do you want an example?

anonymous · Answer

is the answer 1 then ?

anonymous · Answer

would it be: a= 10/11 the bottom would be 1-(10/11)

solomonzelman · Answer

lets do the first series, ok?

anonymous · Answer

yes, thank you.

solomonzelman · Answer

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \left(\frac{2 }{11}\right)^n}$

$\large\color{black}{ \displaystyle a_1=2/11}$

So the sum is:
$\large\color{black}{ \displaystyle S=\frac{a_1 }{1-r}=\frac{\frac{2}{11}}{1-\frac{2}{11}}}$

can you simplify the sum further?

anonymous · Answer

so it comes to be (2/11)/(9/11)

anonymous · Answer

or 2/9

solomonzelman · Answer

yes, and that would simplify to?

solomonzelman · Answer

yes, 2/9 good

solomonzelman · Answer

Ok, can you do

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \left(\frac{8 }{11}\right)^n}$

solomonzelman · Answer

what would this sum be?

anonymous · Answer

8/3 ?

solomonzelman · Answer

yes, very nice

solomonzelman · Answer

So,

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \left(\frac{ 2 }{11}\right)^n+\sum_{ n=1 }^{ \infty } ~ \left(\frac{ 8 }{11}\right)^n}=\frac{2}{9}+\frac{8}{3}$

solomonzelman · Answer

questions?

anonymous · Answer

no, thank you so much!! you have no idea how much clearer this is!
Would you mind helping me with one last question? its different and a bit scary looking

solomonzelman · Answer

Yes, sure...

anonymous · Answer

thank you.. ill try to type it with the equation function option so its understandable

solomonzelman · Answer

or you can draw it if you want to:)

anonymous · Answer

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