Question

Consider the series 1/4 1/16 1/64 1/256 which expression defines sn

Answer 1

This is a geometric progression

• Common ratio=1/16÷1/4 =1/4
• first term =a=1/4

So

The general formUla is

$\\ \rm\Rrightarrow a(n)=ar^{n-1}$

$\\ \rm\Rrightarrow a(n)=\dfrac{1}{4}\left(\dfrac{1}{4}\right)^{n-1}$

Answer 2

Answer:

$\displaystyle \lim_{n \to \infty} \dfrac{1}{3}\left(1-\left(\dfrac{1}{4}\right)^n\right)$

Step-by-step explanation:

Series: the sum of the elements of a sequence.

Therefore, as the numbers have been defined as a series and we need to find $S_n$:

$\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+...$

First determine if the sequence is arithmetic or geometric.

If it is an arithmetic sequence, there will be a common difference between consecutive terms.

if it is a geometric sequence, there will be a common ratio between consecutive terms.

From inspection of the terms, we can see that there is a common ratio of 1/4, as each term is the previous term multiplied by 1/4, so it is a geometric series.

Sum of the first n terms of a geometric series:

$S_n=\dfrac{a(1-r^n)}{1-r}$

Given:

$a=\dfrac{1}{4}$

$r=\dfrac{1}{4}$

Substitute the values of a and r into the formula:

$\implies S_n=\dfrac{\frac{1}{4}\left(1-\left(\frac{1}{4}\right)^n\right)}{1-\frac{1}{4}}$

$\implies S_n=\dfrac{\frac{1}{4}\left(1-\left(\frac{1}{4}\right)^n\right)}{\frac{3}{4}}$

$\implies S_n=\dfrac{1}{3}\left(1-\left(\dfrac{1}{4}\right)^n\right)$

Therefore:

$\displaystyle \lim_{n \to \infty} \dfrac{1}{3}\left(1-\left(\dfrac{1}{4}\right)^n\right)$