Answer: Either

By induction, let the statement be \(H_k: \sum_{n=1}^{k} n^3 = \frac14 k^2(k+1)^2\).

For \(k=1\), \(1^3=1\) and \(\frac14\cdot 1^2\cdot 2^2=1\), so the result is true for \(k=1\).

Assume it is true for \(k\). Then

\(\sum_{n=1}^{k+1} n^3 = \sum_{n=1}^{k} n^3 + (k+1)^3 = \frac14 k^2(k+1)^2 + (k+1)^3\)

\(= (k+1)^2\left(\frac14 k^2 + k + 1\right) = \frac14 (k+1)^2(k^2+4k+4) = \frac14 (k+1)^2(k+2)^2\).

So the formula holds for \(k+1\), and by induction it is true for all \(N\ge 1\).

Using \(\sum_{n=1}^{N} n^2 = \frac16 N(N+1)(2N+1)\),

\(\sum_{n=1}^{N}(20n^3+36n^2)=20\sum_{n=1}^{N}n^3+36\sum_{n=1}^{N}n^2\)

\(=20\cdot \frac14 N^2(N+1)^2 + 36\cdot \frac16 N(N+1)(2N+1)\)

\(=5N^2(N+1)^2+6N(N+1)(2N+1)\)

\(=N(N+1)\bigl(5N(N+1)+6(2N+1)\bigr)=N(N+1)(5N^2+17N+6)\)

\(=N(N+1)(N+3)(5N+2)\).

Now let

\(S_N=\sum_{n=1}^{N}(20n^3+36n^2+\mu n)\).

Then

\(S_N=N(N+1)(5N^2+17N+6)+\mu\sum_{n=1}^{N}n\)

\(=N(N+1)(5N^2+17N+6)+\frac{\mu}{2}N(N+1)\)

\(=N(N+1)\left(5N^2+17N+6+\frac{\mu}{2}\right).\)

For this to be of the form \(N^2(N+1)(aN+b)\), the bracket must have no constant term, so

\(6+\frac{\mu}{2}=0\Rightarrow \mu=-12.\)

Then

\(S_N=N(N+1)(5N^2+17N)=N^2(N+1)(5N+17),\)

so \(a=5\) and \(b=17\).

Also, with \(\mu=-12\),

\(N^{-4}S_N = \frac{N^2(N+1)(5N+17)}{N^4} = \left(1+\frac1N\right)\left(5+\frac{17}{N}\right)=5+\frac{22}{N}+\frac{17}{N^2}.\)

Since \(\frac{17}{N^2} \gt 0\), we have \(5+\frac{22}{N} \lt N^{-4}S_N\). Also, for \(N\ge 18\), \(17/N^2 \lt 1/N\), so

\(N^{-4}S_N \lt 5+\frac{22}{N}+\frac1N = 5+\frac{23}{N}.\)

Hence, for all \(N\ge 18\), \(5+\frac{22}{N} \lt N^{-4}S_N \lt 5+\frac{23}{N}.\)

Either

Let \(H_k\) be the statement \(\sum_{n=1}^{k} n^3 = \frac14 k^2(k+1)^2\).

Base case: when \(k=1\),

\(\sum_{n=1}^{1} n^3 = 1\), and \(\frac14\cdot 1^2\cdot 2^2=1\), so \(H_1\) is true.

Inductive step: assume \(H_k\) is true for some \(k\ge 1\). Then

\(\sum_{n=1}^{k+1} n^3 = \sum_{n=1}^{k} n^3 + (k+1)^3\)

\(= \frac14 k^2(k+1)^2 + (k+1)^3\)

\(= (k+1)^2\left(\frac14 k^2 + k + 1\right)\)

\(= \frac14 (k+1)^2(k^2+4k+4)\)

\(= \frac14 (k+1)^2(k+2)^2.\)

So \(H_{k+1}\) is true. Therefore, by induction,

\(\sum_{n=1}^{N} n^3 = \frac14 N^2(N+1)^2\)

for all positive integers \(N\).

Now use \(\sum_{n=1}^{N} n^2 = \frac16 N(N+1)(2N+1)\). Then

\(\sum_{n=1}^{N}(20n^3+36n^2)=20\sum_{n=1}^{N}n^3+36\sum_{n=1}^{N}n^2\)

\(=20\cdot \frac14 N^2(N+1)^2 + 36\cdot \frac16 N(N+1)(2N+1)\)

\(=5N^2(N+1)^2+6N(N+1)(2N+1)\)

\(=N(N+1)\bigl(5N(N+1)+6(2N+1)\bigr)\)

\(=N(N+1)(5N^2+17N+6)\)

\(=N(N+1)(N+3)(5N+2).\)

Next,

\(S_N=\sum_{n=1}^{N}(20n^3+36n^2+\mu n)\)

\(=N(N+1)(5N^2+17N+6)+\mu\sum_{n=1}^{N}n\)

\(=N(N+1)(5N^2+17N+6)+\frac{\mu}{2}N(N+1)\)

\(=N(N+1)\left(5N^2+17N+6+\frac{\mu}{2}\right).\)

We want \(S_N\) to be of the form \(N^2(N+1)(aN+b)\). Since there is already a factor \(N(N+1)\), the remaining quadratic must factor as \(N(aN+b)\), so its constant term must be zero:

\(6+\frac{\mu}{2}=0\Rightarrow \mu=-12.\)

Then

\(S_N=N(N+1)(5N^2+17N)=N^2(N+1)(5N+17).\)

So \(a=5\) and \(b=17\).

Finally, with \(\mu=-12\),

\(N^{-4}S_N=\frac{N^2(N+1)(5N+17)}{N^4} = \left(1+\frac1N\right)\left(5+\frac{17}{N}\right)\)

\(=5+\frac{22}{N}+\frac{17}{N^2}.\)

This immediately gives \(N^{-4}S_N \gt 5+\frac{22}{N}\).

For \(N\ge 18\), we have \(N\gt 17\), so \(\frac{17}{N^2} \lt \frac1N\). Hence

\(N^{-4}S_N = 5+\frac{22}{N}+\frac{17}{N^2} \lt 5+\frac{22}{N}+\frac1N = 5+\frac{23}{N}.\)

Therefore, for all \(N\ge 18\),

\(5+\frac{22}{N} \lt N^{-4}S_N \lt 5+\frac{23}{N}.\)