Answer:
[tex](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}[/tex]
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 [tex]\frac{1}{3}[/tex]. In the next step, we remove the middle third of the remaining two segments whose lengths are [tex]\frac{1}{3}[/tex], which means that the length of removed parts in this step is [tex]\frac{1}{9} + \frac{1}{9} =\frac{2}{9}[/tex]. In the third step we remove middle thirds of four segments which we got in the previous step whose lengths are [tex]\frac{1}{9}[/tex] each, so, length of removed parts is
[tex]\frac{1}{27} +\frac{1}{27} +\frac{1}{27} +\frac{1}{27}=\frac{4}{27}[/tex]
After the third step, we got eight segments whose lengths are [tex]\frac{1}{27}[/tex] each, which means that we now remove 8 segments whose lengths are [tex]\frac{1}{81}[/tex], or, their total length is [tex]\frac{8}{81}[/tex].
The conclusion is that the first four terms of a series that represents the total length of the segments removed after four steps are
[tex]\frac{1}{3} , \frac{2}{9} , \frac{4}{27} , \frac{8}{81}[/tex]
(b) According to the previous part, we can conclude that the nth term of this sequence is
[tex]a_n=\frac{2^n}{3^{n+1}}[/tex]
and a summation with an nth term that represents the amount removed if this pattern were continued infinitely is
[tex]\sum_{n=0}^{\infty} \frac{2^n}{3^{n+1}}[/tex]
(c) We can notice that the previous sum is actually the geometric progression, where the first term is [tex]\frac{1}{3}[/tex] and the quotient is [tex]\frac{2}{3}[/tex], so, the required sum is
[tex]\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[/tex]
The conclusion is that this infinite series converges.
