1.4 Calculate \( p \) if \( \sum_{k=1}^{\infty} 27 p^{k}=\sum_{t=2}^{5}\left(1-t^{2}\right) \)

To solve for \( p \), we first simplify the right-hand side of the equation. The expression \( \sum_{t=2}^{5}(1-t^{2}) \) can be computed by evaluating each term: \[ 1 - 2^2 = 1 - 4 = -3 \] \[ 1 - 3^2 = 1 - 9 = -8 \] \[ 1 - 4^2 = 1 - 16 = -15 \] \[ 1 - 5^2 = 1 - 25 = -24 \] Now, summing these values gives: \[ -3 + -8 + -15 + -24 = -50 \] Next, we rewrite the left-hand side sum, \( \sum_{k=1}^{\infty} 27 p^k \). This is a geometric series with the first term \( 27p \) and common ratio \( p \): \[ \sum_{k=1}^{\infty} 27 p^k = 27p \cdot \frac{1}{1-p} = \frac{27p}{1-p} \quad \text{for } |p| < 1 \] Setting the left side equal to the right side: \[ \frac{27p}{1-p} = -50 \] To solve for \( p \), we first cross-multiply: \[ 27p = -50(1 - p) \] \[ 27p = -50 + 50p \] Now, we rearrange to isolate \( p \): \[ 27p - 50p = -50 \] \[ -23p = -50 \] \[ p = \frac{50}{23} \] The solution \( p \) is \( \frac{50}{23} \), but we need to ensure this value satisfies the constraint \( |p| < 1 \). Since \( \frac{50}{23} \approx 2.17 \), it does not satisfy \( |p| < 1 \). Thus, a valid \( p \) under the geometric series condition does not exist in real numbers for this equation.