For m, n > 0 , let α(m, n) = ∫ _ 0 ^ 2 t ^ m (1 + 3t) ^ n dt . If ,...

Given,

$\alpha\left(m, n\right) = {∫}_{0}^{2}{t}^{m}{\left(1 + 3t\right)}^{n}dt$

Now using integration by parts we get,

$⇒ \alpha\left(m, n\right) = {{\left(1 + 3t\right)}^{n} ⋅ \dfrac{{t}^{m + 1}}{m + 1}|}_{0}^{2} - {∫}_{0}^{2}n{\left(1 + 3t\right)}^{n - 1} \times 3 ⋅ \dfrac{{t}^{m + 1}}{m + 1}dt$

$⇒ \left(m + 1\right)\alpha\left(m, n\right) = {{\left(1 + 3t\right)}^{n} ⋅ \left({t}^{m + 1}\right)|}_{0}^{2} - 3n\left(m + 1\right){∫}_{0}^{2}{\left(1 + 3t\right)}^{n - 1} \times {t}^{m + 1}dt$

$⇒ \left(m + 1\right)\alpha\left(m, n\right) = {\left(1 + 3 \times 2\right)}^{n} ⋅ \left({2}^{m + 1}\right) - 3n\left(m + 1\right)\alpha\left(m + 1, n - 1\right)$

$⇒ \left(m + 1\right)\alpha\left(m, n\right) = {7}^{n} \cdot {2}^{m + 1} - 3n\alpha\left(m + 1, n - 1\right)$

$⇒ \left(m + 1\right)\alpha\left(m, n\right) + 3n\alpha\left(m + 1, n - 1\right) = {7}^{n} \cdot {2}^{m + 1}$

Now put $m = 10, n = 6$ in above equation we get,

$11\alpha\left(10, 6\right) + 18\alpha\left(11, 5\right) = {7}^{6} \cdot {2}^{11} = 32 \times {\left(14\right)}^{6}$

Hence, on comparing with $11\alpha\left(10, 6\right) + 18\alpha\left(11, 5\right) = p \times {\left(14\right)}^{6}$ we get,

$⇒ p = 32$