With some rewriting, you get
[tex]\displaystyle \sum_{k=0}^\infty (-1)^k\frac{x^{k+2}}{4^k} = x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k[/tex]
Recall that for |x| < 1, you have
[tex]\displaystyle \frac1{1-x} = \sum_{k=0}^\infty x^k[/tex]
So as long as |-x/4| = |x/4| < 1, or |x| < 4, your series converges to
[tex]\displaystyle x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k = \frac{x^2}{1-\left(-\frac x4\right)} = \frac{x^2}{1+\frac x4} = \boxed{\frac{4x^2}{4+x}}[/tex]
Based on known expressions from Taylor series, the power series [tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex]Taylor series-derived formula of the rational function [tex]\frac{4\cdot x^{2}}{4+x}[/tex].
Taylor series are polynomic approximations used to estimate values both from trascendental and non-trascendental functions. It is commonly used in trigonometric, potential, logarithmic and even rational functions.
In this question we must use series properties and common Taylor series-derived formulas to infer the expression behind the given series. Now we proceed to find the expression:
[tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex]
[tex]x^{2}\cdot \sum\limits_{k = 0}^{\infty} \left(-\frac{x}{4} \right)^{k}[/tex]
[tex]x^{2}\cdot \left(\frac{1}{1+\frac{x}{4} } \right)[/tex]
[tex]\frac{4\cdot x^{2}}{4+x}[/tex]
Based on power and series properties and most common Taylor series- derived formulas, the power series [tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex] represents a Taylor series-derived formula of the rational function [tex]\frac{4\cdot x^{2}}{4+x}[/tex]. [tex]\blacksquare[/tex]
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