Question

If a_a is the greatest term in the sequence \(a_n=\frac{n^3}{n^4+147},n=1,2,3,...,\) then a is equal to______________.

Hint:

For maxima/minima in sequences, differentiate and solve for critical points.

Answer 1

Correct Answer: 5

Step 1: Define \(f(x)\) and find the derivative.
\[ f(x) = \frac{x^3}{x^4 + 147}, \quad f'(x) = \frac{3x^2(x^4 + 147) - x^3(4x^3)}{(x^4 + 147)^2}. \] Step 2: Solve \(f'(x) = 0\).
\[ 3x^6 + 441x^2 - 4x^6 = 0 \implies x^2(441 - x^4) = 0. \] \[ x = \sqrt[4]{441} \approx 3.5. \] Step 3: Find \(f(3.5)\).
\[ f(3.5) = \frac{(3.5)^3}{(3.5)^4 + 147} \approx 0.158. \] Final Answer: The greatest term is \(0.158\).
\[ a = 5 \]