Question

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

Answer 1

Correct Answer: 9

To solve \(8 = 3 + \frac{1}{4}(3 + p) + \frac{1}{4^2}(3 + 2p) + \frac{1}{4^3}(3 + 3p) + \ldots\), we identify a geometric series with a common ratio of \( \frac{1}{4} \). Let the series be \( S \). The equation can be rewritten as \( S = 3 + \sum_{n=1}^{\infty} \frac{1}{4^n}(3+np) \). Separating this into two series yields: \[ S = 3 + \sum_{n=1}^{\infty} \frac{3}{4^n} + \sum_{n=1}^{\infty} \frac{np}{4^n} \] The first series, \( \sum_{n=1}^{\infty} \frac{3}{4^n} \), is a geometric series with first term \( a = \frac{3}{4} \) and common ratio \( r = \frac{1}{4} \). Its sum is: \[ \sum_{n=1}^{\infty} \frac{3}{4^n} = \frac{\frac{3}{4}}{1 - \frac{1}{4}} = \frac{3}{4} \times \frac{4}{3} = 1 \] The second series, \( \sum_{n=1}^{\infty} \frac{np}{4^n} \), is an arithmetico-geometric series. Its sum is calculated using the formula \( \sum_{n=1}^{\infty} \frac{n}{r^n} = \frac{r}{(1-r)^2} \) with \( x = \frac{1}{4} \). Thus, the sum is: \[ \sum_{n=1}^{\infty} \frac{np}{4^n} = \frac{\frac{1}{4}}{(1 - \frac{1}{4})^2} \times p = \frac{\frac{1}{4}}{\frac{9}{16}} \times p = \frac{4}{9}p \] Substituting these sums back into the equation for \( S \): \[ S = 3 + 1 + \frac{4}{9}p = 8 \] This simplifies to: \[ 4 + \frac{4}{9}p = 8 \] \[ \frac{4}{9}p = 4 \] Solving for \( p \): \[ p = 9 \] The value of \( p \) is 9, which lies within the given range of \([9,9]\).