If 8=3+(1)/(4)(3+p)+(1)/(4^2)(3+2 p)+(1)/(4^3)(3+3 p)+ , then the value of p is __________…

Question

If $8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2 p)+\frac{1}{4^3}(3+3 p)+\cdots \cdots \infty$, then the value of $p$ is \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 9

Accepted Answer

$8=\frac{3}{1-\frac{1}{4}}+\frac{p\cdot \frac{1}{4}}{{(1-\frac{1}{4})}^{2}}$

$=\frac{a}{1-r}+\frac{dr}{(1-r{)}^{2}}$

$\Rightarrow \frac{4p}{9}=4\Rightarrow p=9$

Solutions · 1