Question

find the maclaurin series for the given function
what is the maclaurin series of ( f(x)=\frac{8}{1 + x} )?
( sum_{k = 0}^{infty}square ) (type an expression using ( k ) and ( x ) as the variables.)

Explanation:

Step1: Recall the geometric series formula

The geometric series is \(\sum_{k = 0}^{\infty}r^{k}=\frac{1}{1 - r}\), for \(|r|\lt1\). In our case, \(r=-x\), so \(\frac{1}{1+x}=\sum_{k = 0}^{\infty}(-x)^{k}=\sum_{k = 0}^{\infty}(- 1)^{k}x^{k}\), for \(|x|\lt1\).

Step2: Multiply by the constant factor

We have \(f(x)=\frac{8}{1 + x}\). Since \(\frac{1}{1+x}=\sum_{k = 0}^{\infty}(-1)^{k}x^{k}\), then \(f(x)=8\times\frac{1}{1+x}\).
By the property of series \(\sum_{k = 0}^{\infty}a\times u_{k}=a\times\sum_{k = 0}^{\infty}u_{k}\) (where \(a = 8\) and \(u_{k}=(-1)^{k}x^{k}\)), we get \(f(x)=\sum_{k = 0}^{\infty}8(-1)^{k}x^{k}\).

Answer:

\(\sum_{k = 0}^{\infty}8(-1)^{k}x^{k}\)