\( \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{4} \) Calculate the value of \( p \) The value of \( p=5 \)

To solve the equation \[ \sum_{k=p}^{\infty} 4 \cdot 3^{2-k} = \frac{2}{4} \] we first simplify the right side: \[ \frac{2}{4} = \frac{1}{2} \] Now, we can rewrite the left side of the equation. The series can be expressed as: \[ \sum_{k=p}^{\infty} 4 \cdot 3^{2-k} = 4 \cdot \sum_{k=p}^{\infty} 3^{2-k} \] Next, we can factor out the constant \( 4 \): \[ = 4 \cdot 3^2 \cdot \sum_{k=p}^{\infty} \left(\frac{1}{3}\right)^{k} \] The series \( \sum_{k=p}^{\infty} \left(\frac{1}{3}\right)^{k} \) is a geometric series with the first term \( a = \left(\frac{1}{3}\right)^{p} \) and the common ratio \( r = \frac{1}{3} \). The sum of an infinite geometric series is given by: \[ S = \frac{a}{1 - r} \] Substituting the values, we have: \[ S = \frac{\left(\frac{1}{3}\right)^{p}}{1 - \frac{1}{3}} = \frac{\left(\frac{1}{3}\right)^{p}}{\frac{2}{3}} = \frac{3}{2} \cdot \left(\frac{1}{3}\right)^{p} = \frac{3^{1-p}}{2} \] Now substituting this back into our equation: \[ 4 \cdot 3^2 \cdot \frac{3^{1-p}}{2} = \frac{1}{2} \] This simplifies to: \[ 4 \cdot 9 \cdot \frac{3^{1-p}}{2} = \frac{1}{2} \] \[ 18 \cdot 3^{1-p} = \frac{1}{2} \] Now, we can isolate \( 3^{1-p} \): \[ 3^{1-p} = \frac{1}{36} \] Taking the logarithm base 3 of both sides: \[ 1 - p = \log_3\left(\frac{1}{36}\right) \] We can express \( \frac{1}{36} \) as \( 3^{-2} \): \[ 1 - p = \log_3(3^{-2}) = -2 \] Thus, we have: \[ 1 - p = -2 \implies p = 3 \] Therefore, the value of \( p \) is \[ \boxed{3} \]