Derivatives of the Infinite Geometric Progression

Calculus Level pending

Use the fact that

\[\frac{1}{2 - x} = \sum_{n = 0}^{\infty}\frac{x^n}{2^{n + 1}} ~\text{ for }~ \left |x \right |<1 \]

to determine a power series for \(\displaystyle\frac{1}{(2 - x)^2} ~\text{ for }~ \left |x \right |<1 \).

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