Recall that for |x| < 1, we have
[tex]\displaystyle \frac1{1-x} = \sum_{n=0}^\infty x^n[/tex]
Differentiating both sides gives
[tex]\displaystyle \frac1{(1-x)^2} = \sum_{n=0}^\infty n x^{n-1} = \sum_{n=0}^\infty (n+1) x^n[/tex]
so that the Maclaurin expansion of the given function is
[tex]\displaystyle \frac5{1-x} = \boxed{\sum_{n=0}^\infty 5(n+1) x^n} = 5 + 10x + 15x^2 + 20x^3 + \cdots[/tex]
