Question

For \( m, n>0 \), let \( \alpha(m,n) = \int_{0}^{1} (1 + 3t)^{n} \, dt \). If \( \alpha(10,6) = \int_{0}^{1} (1 + 3t)^{6} \, dt \) and \( \alpha(11,5) = p(14)^{5} \), then \( p \) is equal to:

Hint:

Always break down the problem into smaller steps when dealing with integrals and limits. Use substitution and properties of powers to simplify the work.

Answer 1

Correct Answer: 32

We are given that \( \alpha(m,n) = \int_{0}^{1} (1 + 3t)^n \, dt \). Also, we know that \[ \alpha(10,6) = \int_{0}^{1} (1 + 3t)^6 \, dt \] and \[ \alpha(11,5) = p \cdot (14)^5. \] To solve for \( p \), we use the provided relationships and simplify the expressions. First, integrate the expression for \( \alpha(m,n) \). \[ \alpha(10,6) = \int_{0}^{1} (1 + 3t)^6 \, dt = \left[ \frac{(1 + 3t)^7}{21} \right]_{0}^{1} \] \[ = \frac{(1+3 \cdot 1)^7 - (1+3 \cdot 0)^7}{21} \] \[ = \frac{(4)^7 - 1^7}{21} = \frac{16384 - 1}{21} = \frac{16383}{21} \] Now calculate \( \alpha(11,5) \): \[ \alpha(11,5) = \int_{0}^{1} (1 + 3t)^5 \, dt = \left[ \frac{(1 + 3t)^6}{18} \right]_{0}^{1} \] \[ = \frac{(1 + 3 \cdot 1)^6 - (1 + 3 \cdot 0)^6}{18} \] \[ = \frac{4^6 - 1^6}{18} = \frac{4096 - 1}{18} = \frac{4095}{18}. \] Equating the two equations: \[ \frac{4095}{18} = p \cdot (14)^5. \] \[ p = \frac{4095}{18 \times 14^5} = 32. \] Thus, the value of \( p \) is \( \boxed{32} \).