Recall that
[tex]\displaystyle\frac1{1-3x}=\sum_{n\ge0}(3x)^n[/tex]
for [tex]|3x|<1[/tex]. Notice that
[tex]\dfrac{\mathrm d}{\mathrm dx}\dfrac1{1-3x}=\dfrac3{(1-3x)^2}[/tex]
so that we have
[tex]\displaystyle\frac{15}{(1-3x)^2}=15\frac{\mathrm d}{\mathrm dx}\sum_{n\ge0}(3x)^n[/tex]
[tex]\implies\displaystyle\frac{15}{(1-3x)^2}=15\sum_{n\ge1}n(3x)^{n-1}[/tex]
(again, for [tex]|3x|<1[/tex])
