If $8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2p)+\frac{1}{4^3}(3+3p)+\dots <!--\; \dots \; -->\mathrm{\infty <!--\; \infty \; -->}$, then the value of p is ____

The correct answer is p = 9.

We are given the infinite series equation:

8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2p)+\frac{1}{4^3}(3+3p)+\dots \infty

Step 1: Write the series in summation form.

The terms after the leading 3 follow a clear pattern. The n-th term (starting from n = 1) is:

\frac{1}{4^n}(3+np)

So the equation becomes:

8=3+\sum _{n=1}^{\infty }\frac{1}{4^n}(3+np)

Step 2: Split the summation into two separate series.

8=3+3\sum _{n=1}^{\infty }{\left(\frac{1}{4}\right)}^n+p\sum _{n=1}^{\infty }n{\left(\frac{1}{4}\right)}^n

Step 3: Evaluate the first summation — a standard geometric series.

Using the formula \sum _{n=1}^{\infty }{x}^n=\frac{x}{1-x} with x=\frac{1}{4}:

\sum _{n=1}^{\infty }{\left(\frac{1}{4}\right)}^n=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}

Step 4: Evaluate the second summation — an arithmetico-geometric series.

Using the standard result \sum _{n=1}^{\infty }n{x}^n=\frac{x}{{(1-x)}^2} with x=\frac{1}{4}:

\sum _{n=1}^{\infty }n{\left(\frac{1}{4}\right)}^n=\frac{\frac{1}{4}}{{(1-\frac{1}{4})}^2}=\frac{\frac{1}{4}}{{\left(\frac{3}{4}\right)}^2}=\frac{\frac{1}{4}}{\frac{9}{16}}=\frac{4}{9}

Step 5: Substitute both results back into the equation.

8=3+3\times \frac{1}{3}+p\times \frac{4}{9}

8=3+1+\frac{4p}{9}

8=4+\frac{4p}{9}

Step 6: Solve for p.

8-4=\frac{4p}{9}

4=\frac{4p}{9}

p=\frac{4\times 9}{4}=9

Therefore, the value of p = 9.