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

 To determine where the sequence \(a_n=\frac{n^3}{n^4+147}\) attains its greatest value, we need to differentiate the expression with respect to \(n\) and find its critical points. Define the function \(f(n)=\frac{n^3}{n^4+147}\).

The derivative \(f'(n)\) is computed using the quotient rule: if \(u(n)=n^3\) and \(v(n)=n^4+147\), then \(f(n)=\frac{u(n)}{v(n)}\) and:

\(f'(n)=\frac{v(n)u'(n)-u(n)v'(n)}{(v(n))^2}=\frac{(n^4+147)(3n^2)-(n^3)(4n^3)}{(n^4+147)^2}\)

Simplify to find the critical points:

\(f'(n)=\frac{3n^6+441n^2-4n^6}{(n^4+147)^2}=\frac{-n^6+441n^2}{(n^4+147)^2}\)

Setting \(f'(n)=0\) to find critical points, we solve:

\(-n^6+441n^2=0 \Rightarrow n^2(n^4-441)=0 \Rightarrow n^4=441\)

Solving \(n^4=441\), we find \(n^2=\sqrt{441}=21\), thus \(n=\sqrt{21}\). However, since \(n\) must be an integer, we approximate and test values around \(\sqrt{21}\): \(n=4\) and \(n=5\).

Evaluate these in \(f(n)\):

\(f(4)=\frac{4^3}{4^4+147}=\frac{64}{256+147}=\frac{64}{403}\approx 0.159\)

\(f(5)=\frac{5^3}{5^4+147}=\frac{125}{625+147}=\frac{125}{772}\approx 0.162\)

Since \(f(5)>f(4)\), \(a_n\) has its maximum at \(n=5\).

Verify \(n=5\) is within the specified range \(5,5\). Therefore, the greatest term is at \(n=a=5\).

Conclusion: \(a=5\)