If 8 = 3 + (1)/(4)(3 + p) + 1 4 ^ 2 (3 + 2p) + 1 4 ^ 3 (3 + 3p) + ∞...

Given that $3 + \dfrac{1}{4}\left(3 + p\right) + \dfrac{1}{{4}^{2}}\left(3 + 2p\right) + \ldots .$ to $\infty = 8$

Let, $S = 3 + \dfrac{1}{4}\left(3 + p\right) + \dfrac{1}{{4}^{2}}\left(3 + 2p\right) + . . . \infty$

$⇒ S = \left(3 + \dfrac{3}{4} + \dfrac{3}{{4}^{2}} + . . .\right) + \left(\dfrac{p}{4} + \dfrac{2p}{{4}^{2}} + . . .\right)$

We know that, ${S}_{\infty} = a + ar + {ar}^{2} + . . . \infty = \dfrac{a}{1 - r}$

Let, ${S}_{1} = 3 + \dfrac{3}{4} + \dfrac{3}{{4}^{2}} + . . . .$

$⇒ {S}_{1} = 3\left(1 + \dfrac{1}{4} + \dfrac{1}{{4}^{2}} + . . . .\right)$

$⇒ {S}_{1} = 3\left(\dfrac{1}{1 - \dfrac{1}{4}}\right) = 4$

And, ${S}_{2} = \dfrac{p}{4} + \dfrac{2p}{{4}^{2}} + \dfrac{3p}{{4}^{3}} + . . . .$ 

$⇒ \dfrac{{S}_{2}}{4} = \dfrac{p}{{4}^{2}} + \dfrac{2p}{{4}^{3}} + \dfrac{3p}{{4}^{4}} + . . . .$

$⇒ {S}_{2} - \dfrac{{S}_{2}}{4} = \dfrac{p}{4} + \dfrac{p}{{4}^{2}} + \dfrac{p}{{4}^{3}} + . . . . .$

$⇒ \dfrac{3{S}_{2}}{4} = p\left(\dfrac{1}{4} + \dfrac{1}{{4}^{2}} + \dfrac{1}{{4}^{3}} + . . . . .\right)$

$⇒ \dfrac{3{S}_{2}}{4} = p\left(\dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{4}}\right) = \dfrac{p}{3}$

$⇒ {S}_{2} = \dfrac{4p}{9}$

So, $S = {S}_{1} + {S}_{2} = 8$

$⇒ 4 + \dfrac{4p}{9} = 8$

$⇒ \dfrac{4p}{9} = 4$

$⇒ p = 9$