Certainly! Let's solve the given question step-by-step.
We are given that [tex]\( a_k = \frac{1}{k(k+1)} \)[/tex] for [tex]\( k = 1, 2, 3, \ldots \)[/tex].
Additionally, we need to find the following fractions in simplified form for variable [tex]\( n \)[/tex]:
1) [tex]\(\frac{n}{n+1}\)[/tex]
2) [tex]\(\frac{n^2}{(n+1)^2}\)[/tex]
3) [tex]\(\frac{n^4}{(n+1)^4}\)[/tex]
4) [tex]\(\frac{n^6}{(n+1)^6}\)[/tex]
### 1. Simplifying [tex]\(\frac{n}{n+1}\)[/tex]:
This fraction is already in its simplest form.
### Result:
[tex]\[\frac{n}{n+1}\][/tex]
### 2. Simplifying [tex]\(\frac{n^2}{(n+1)^2}\)[/tex]:
This fraction is also in its simplest form since [tex]\(n\)[/tex] and [tex]\(n+1\)[/tex] are co-prime (they have no common factors other than 1).
### Result:
[tex]\[\frac{n^2}{(n+1)^2}\][/tex]
### 3. Simplifying [tex]\(\frac{n^4}{(n+1)^4}\)[/tex]:
Again, this fraction cannot be simplified further because [tex]\(n\)[/tex] and [tex]\(n+1\)[/tex] are co-prime.
### Result:
[tex]\[\frac{n^4}{(n+1)^4}\][/tex]
### 4. Simplifying [tex]\(\frac{n^6}{(n+1)^6}\)[/tex]:
This fraction is already simplified for the same reason as above: [tex]\(n\)[/tex] and [tex]\(n+1\)[/tex] are co-prime.
### Result:
[tex]\[\frac{n^6}{(n+1)^6}\][/tex]
To summarize, the fractions [tex]\(\frac{n}{n+1}\)[/tex], [tex]\(\frac{n^2}{(n+1)^2}\)[/tex], [tex]\(\frac{n^4}{(n+1)^4}\)[/tex], and [tex]\(\frac{n^6}{(n+1)^6}\)[/tex] are already in their simplest forms and cannot be further reduced. The answers are:
1. [tex]\[\frac{n}{n+1}\][/tex]
2. [tex]\[\frac{n^2}{(n+1)^2}\][/tex]
3. [tex]\[\frac{n^4}{(n+1)^4}\][/tex]
4. [tex]\[\frac{n^6}{(n+1)^6}\][/tex]
