Question

Find the value of a given
$3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2p)+\frac{1}{4^3}(3+3p)+...=?$

Answer 1

$3+\frac{1}{4^{}}(3+p)+\frac{1}{4^2}(3+2p)+\frac{1}{4^3}(3+3p)+....$

$S=3+\frac{3}{4^{}}+\frac{3}{4^2}+\frac{3}{4^3}+.....\frac{p}{4^{}}+\frac{p}{4^2}+\frac{p}{4^3}+....$

$S=3(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....)+p(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+.....)=3\left(\frac{1}{1-\frac{1}{4}}\right)+p(\frac{1}{4}+\frac{2}{4^2}+....)$

$S=3\times \frac{4}{3}+p(\frac{1}{4}+\frac{2}{4^2}+....)$

$S-4=p(\frac{1}{4}+\frac{2}{4^2}+....)..\left(I\right)$

$\frac{S-4}{4}=p(\frac{2}{4^2}+\frac{3}{4^3}....)....\left(II\right)$

$\left(II\right)-\left(I\right)$

$-3\frac{S-4}{4}=-\frac{p}{4}-\frac{p}{4^2}+\frac{p}{4^3}+...=p\left(\frac{1/4}{1-1/4}\right)\frac{3}{4}(S-4)=\frac{p}{3}$

$\Rightarrow S-4=\frac{4p}{9}\Rightarrow S=\frac{4p+36}{9}$