If $a_{\alpha}$ is the greatest term in the sequence $a_{n}=\frac{n^{3}}{n^{4}+147}, n=1,2,3, \ldots

Question

If $a_{\alpha}$ is the greatest term in the sequence $a_{n}=\frac{n^{3}}{n^{4}+147}, n=1,2,3, \ldots .$. , then $\alpha$ is equal to

Professor PC Thomas Classes · Answer

(5) : We have, $a_{n}=\frac{n^{3}}{n^{4}+147}$ Differentiate both sides w.r.t. $n$, we get\ $\Rightarrow \quad a_{n}^{\prime}=\frac{\left(n^{4}+147ight) 3 n^{2}-n^{3} \cdot 4 n^{3}}{\left(n^{4}+147ight)^{2}}$\ Put, $a_{n}^{\prime}=0$\ $\Rightarrow \frac{n^{2}\left\{3\left(n^{4}+147ight)-4 n^{4}ight\}}{\left(n^{4}+147ight)^{2}}=0$\ $\Rightarrow \quad 3 n^{4}+441-4 n^{4}=0 \Rightarrow n^{4}=441$\ $\Rightarrow n^{2}=21 \Rightarrow n=\sqrt{21}$\ $\Rightarrow 4<\sqrt{21}<5 	herefore a_{5}>a_{4}$\ So, $\alpha=5$\