From the odd-degree terms, take out one copy and rewrite the series as
[tex]1+6x+x^2+6x^3+\cdots=(1+x+x^2+x^3+\cdots)+5x+5x^3+\cdots[/tex]
[tex]1+6x+x^2+6x^3+\cdots=(1+x+x^2+x^3+\cdots)+5x(1+x^2+\cdots)[/tex]
Then if |x| < 1, we can condense this to
[tex]\displaystyle\sum_{n=0}^\infty x^n+5x\sum_{n=0}^\infty x^{2n}=\frac1{1-x}+\frac{5x}{1-x^2}=\frac{1+6x}{1-x^2}[/tex]
Since the series we invoked here converge on -1 < x < 1, so does this one.
The explicit formula of the function f(x) is [tex]f(x) = \frac{1 + x + 5x}{1-x^2}[/tex]
The function definition is given as:
[tex]f(x) = 1 + 6x + x^2 + 6x^3 + x^4 + ...[/tex]
Expand the terms of the expression
[tex]f(x) = 1 + 5x + x + x^2 + 5x^3 + x^3 + x^4 + ...[/tex]
Split
[tex]f(x) = (1 + x + x^2 +x^3 + .....) + 5x + 5x^3 + .. ...[/tex]
Factor out 5x
[tex]f(x) = (1 + x + x^2 +x^3 + .....) + 5x(1 + x^2) + .. ...[/tex]
Express 1 as x^0
[tex]f(x) = (x^0 + x + x^2 +x^3 + .....) + 5x(1 + x^2) + .. ...[/tex]
Express x as x^1
[tex]f(x) = (x^0 + x^1 + x^2 +x^3 + .....) + 5x(1 + x^2) + .. ...[/tex]
Also, we have:
[tex]f(x) = (x^0 + x^1 + x^2 +x^3 + .....) + 5x(x^0 + x^2) + .. ...[/tex]
Rewrite the series using the summation symbol
[tex]f(x) = \sum\limits^{\infty}_{n=0}x^n+ 5x\sum\limits^{\infty}_{n=0}x^{2n}[/tex]
The sum to infinity of a geometric progression is:
[tex]S_{\infty} = \frac{a}{1- r}[/tex]
Where:
a represents the first term, and r represents the common ratio
Using the above formula, we have:
[tex]\sum\limits^{\infty}_{n=0}x^n = \frac{1}{1 - x}[/tex]
[tex]5x\sum\limits^{\infty}_{n=0}x^{2n} = 5x * \frac{1}{1 - x^2} = \frac{5x}{1-x^2}[/tex]
So, we have:
[tex]f(x) = \frac{1}{1-x}+ \frac{5x}{1-x^2}[/tex]
Take the LCM
[tex]f(x) = \frac{1 + x + 5x}{1-x^2}[/tex]
Evaluate the like terms
[tex]f(x) = \frac{1 + 6x}{1-x^2}[/tex]
Hence, the explicit formula of the function f(x) is [tex]f(x) = \frac{1 + x + 5x}{1-x^2}[/tex]
Read more about geometric series at:
https://brainly.com/question/12563588
