In the expansion of $\left(a \sqrt{a} + \dfrac{1}{a^4}\right)^n$, the coefficient in the second term exceeds by 44 the coefficient in the first term. Find $n$.
In the expansion of $\left(a \sqrt{a} + \dfrac{1}{a^4}\right)^n$,
coefficient in first term $= C^n_1$; $\;$ coefficient in second term $= C^n_2$
As per question,
$C^n_2 = C^n_1 + 44$
i.e. $\;$ $\dfrac{n!}{2!\left(n-2\right)!} = \dfrac{n!}{\left(n-1\right)!} + 44$
i.e. $\;$ $\dfrac{n \left(n-1\right) \left(n-2\right)!}{2\left(n-2\right)!} = \dfrac{n \left(n-1\right)!}{\left(n-1\right)!} + 44$
i.e. $\;$ $\dfrac{n \left(n-1\right)}{2} = n + 44$
i.e. $\;$ $n \left(\dfrac{n-1}{2} - 1\right) = 44$
i.e. $\;$ $\dfrac{n \left(n-3\right)}{2} = 44$
i.e. $\;$ $n^2 - 3n - 88 = 0$
i.e. $\;$ $\left(n-11\right)\left(n+8\right) = 0$
i.e. $\;$ $n = 11$ $\;$ or $\;$ $n = -8$
$\because \;$ $n$ cannot be negative as it is the total number of objects
$\therefore \;$ $n = 11$