Question

below (Stage 0) has a length of one unit. Remove the middle third of this line segment (Stage 1). Then remove the middle third of the remaining two segments (Stage 2). Continue this pattern of removing the middle third of any segments that remain, forever. a. Write the first four terms of a series that represents the total length of the segments removed after four stages (Stages 1, 2, 3, and 4). b. Then write a summation with an nth term that represents the amount removed if this pattern were continued infinitely. c. Determine if this infinite series converges or diverges, and clearly explain how you know

Answer 1

Answer:

$(a) \frac{1}{3} , \frac{2}{9} \frac{4}{27} , \frac{8}{81} ; (b) a_n = \frac{2^n}{3^{n+1}} , \sum_{n=0}^{\infty} \frac{2^n}{3^{n+1}}; (c) \text{it converges, its sum is equal to 1}$

Step-by-step explanation:

(a) In the first step, we remove the middle third of a line segment whose length is 1, which means that the length of the removed middle third is $\frac{1}{3}$. In the next step, we remove the middle third of the remaining two segments whose lengths are $\frac{1}{3}$, which means that the length of removed parts in this step is $\frac{1}{9} + \frac{1}{9} =\frac{2}{9}$. In the third step we remove middle thirds of four segments which we got in the previous step whose lengths are $\frac{1}{9}$ each, so, length of removed parts is

$\frac{1}{27} +\frac{1}{27} +\frac{1}{27} +\frac{1}{27}=\frac{4}{27}$

After the third step, we got eight segments whose lengths are $\frac{1}{27}$ each, which means that we now remove 8 segments whose lengths are $\frac{1}{81}$, or, their total length is $\frac{8}{81}$.

The conclusion is that the first four terms of a series that represents the total length of the segments removed after four steps are

$\frac{1}{3} , \frac{2}{9} , \frac{4}{27} , \frac{8}{81}$

(b) According to the previous part, we can conclude that the nth term of this sequence is

$a_n=\frac{2^n}{3^{n+1}}$

and a summation with an nth term that represents the amount removed if this pattern were continued infinitely is

$\sum_{n=0}^{\infty} \frac{2^n}{3^{n+1}}$

(c) We can notice that the previous sum is actually the geometric progression, where the first term is $\frac{1}{3}$ and the quotient is $\frac{2}{3}$, so, the required sum is

$\sum_{n=0}^{\infty} \frac{2^n}{3^{n+1}} = \frac{1}{3} +\frac{2}{9} +\frac{4}{27} + ...=\frac{1}{3}(\frac{1}{1-\frac{2}{3} } ) =\frac{1}{3} *\frac{3}{1} =1$

The conclusion is that this infinite series converges.