Maths Help (1 Viewer)

Farhanthestudent005 · Dec 11, 2021

How to find the limiting sum of the series: 81+27+9....

I kind of forgot..will appreciate if someone assisted me here since I have an exam in less than 2 days, thanks

idkkdi · Dec 11, 2021

Farhanthestudent005 said:
How to find the limiting sum of the series: 81+27+9....

I kind of forgot..will appreciate if someone assisted me here since I have an exam in less than 2 days, thanks

plug numbers into the limiting sum formula.

jimmysmith560 · Dec 11, 2021

As mentioned above, the limiting sum is determined by using the limiting sum formula:

I believe the answer would therefore be as follows:

I hope this helps!

5uckerberg · Dec 12, 2021

Farhanthestudent005 said:
How to find the limiting sum of the series: 81+27+9....

I kind of forgot..will appreciate if someone assisted me here since I have an exam in less than 2 days, thanks

A methodical way of approaching this question would be to ask oneself what is the sum of the geometric series. One will learn that the formula is $S_{n}=\frac{a(1-r^{n})}{1-r}$ which is simply $S_{n}=\frac{a(r^{n}-1)}{r-1}$ but with both the numerator and the denominator multiplied by $-1$
Now, as one can see as $n\rightarrow\infty$ the sum becomes
$S_{\infty}=\frac{a}{1-r}$ noting that r is between -1 and 1.
Do you notice why the a stays in place? The reason is that a is the first term in the series and that r represents the rate of the series and n is the number of terms. A simple example would be to investigate this series $1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n-1}$ now does one notice that $1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n-1}$ is just $\frac{1-x^{n}}{1-x}$ . To know where the a came from just multiply both sides by a and you are done.
Rest is plug and chug.

Maths Help (1 Viewer)

Farhanthestudent005

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idkkdi

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jimmysmith560

Le Phénix Trilingue

5uckerberg

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