If a _ α is the greatest term in the sequence a _ n = n ^ 3 n ^ 4 +...

Let,

$y = \dfrac{{x}^{3}}{{x}^{4} + 147} = f\left(x\right)$

We know that, for increasing function $\dfrac{dy}{dx} > 0$

So, differentiating the given function $f\left(x\right) = \dfrac{{x}^{3}}{{x}^{4} + 147}$ we get,

$\dfrac{dy}{dx} = \dfrac{- {x}^{2}\left({x}^{4} - 441\right)}{( {x}^{4} + 147{)}^{2}}$

$⇒ \dfrac{- {x}^{2}\left({x}^{4} - 441\right)}{( {x}^{4} + 147{)}^{2}} > 0$

$⇒ \dfrac{{x}^{2}\left({x}^{4} - 441\right)}{( {x}^{4} + 147{)}^{2}} < 0$

$⇒ {x}^{4} < 441$

Now for maxima/minima $\dfrac{dy}{dx} = 0$

$⇒ {x}^{4} = 441$

$⇒ x = \alpha, 4 < \alpha < 5$

$⇒$Maximum value of $f$\left(x\right)$is at$x = 4$or$x = 5$</p><p>$f\left(4\right) = \dfrac{64}{403}, f\left(5\right) = \dfrac{125}{772}$</p><p>$∵ f\left(5\right) > f\left(4\right)$</p><p>$⇒ \alpha = 5