Respuesta :

LammettHash
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])

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