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

Question

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

Professor PC Thomas Classes · Answer

Solve $\frac{d}{dx} (\frac{x^3}{x^4+147}) = 0 \Rightarrow x^4 = 3 	imes 147 = 441$. $x = \sqrt{21} \approx 4.5$. Testing integers $n=4, 5$, maximum at $n=5$.