1 4 - 5 4 · 8 + 5 · 9 4 · 8 · 12 - … =

We have,

$\frac{1}{4}-\frac{5}{4·8}+\frac{1·5·9}{4·8·12}-\dots \dots$

$=\frac{1}{4}-\frac{1·\left(1+4\right)}{2!}{\left(\frac{1}{4}\right)}^2+\frac{1·\left(1+4\right)\left(1+2·4\right)}{3!}{\left(\frac{1}{4}\right)}^3-\dots \dots$

$=\left[1\left(\frac{1}{4}\right)-\frac{1·\left(1+4\right)}{2!}{\left(\frac{1}{4}\right)}^2+\frac{1·\left(1+4\right)\left(1+2·4\right)}{3!}{\left(\frac{1}{4}\right)}^3-\dots \dots \right] ...\left(i\right)$

Now, we know that

${\left(1+x\right)}^{-\frac{p}{q}}=1-p\left(\frac{x}{q}\right)+\frac{p\left(p+q\right)}{2!}{\left(\frac{x}{q}\right)}^2-\dots \dots \dots \dots$

Therefore, from $\left(i\right)$, we have

$=\left[1\left(\frac{1}{4}\right)-\frac{1·\left(1+4\right)}{2!}{\left(\frac{1}{4}\right)}^2+\frac{1·\left(1+4\right)\left(1+2·4\right)}{3!}{\left(\frac{1}{4}\right)}^3-\dots \dots \right]$

$=1-\left[1-1\left(\frac{1}{4}\right)+\frac{1·\left(1+4\right)}{2!}{\left(\frac{1}{4}\right)}^2-\frac{1·\left(1+4\right)\left(1+2·4\right)}{3!}{\left(\frac{1}{4}\right)}^3+\dots \dots \right]$

Comparing with the above standard series, we get

$p=1, \frac{x}{q}=\frac{1}{4}, q=4\Rightarrow x=1$

Therefore,

${\left(1+x\right)}^{-\frac{p}{q}}={\left(1+1\right)}^{-\frac{1}{4}}=2^{-\frac{1}{4}}$

Therefore, required sum is

$=\left(1-2^{-\frac{1}{4}}\right)$

$=\left(\frac{2^{\frac{1}{4}}-1}{2^{{\displaystyle \frac{1}{4}}}}\right)$