Answer: (a) Let \(H_n\) be the statement

\(\displaystyle \sum_{r=1}^{n}(3r^{5}+r^{3})=\frac12 n^{3}(n+1)^{3}.\)

For \(n=1\),

\(\displaystyle \sum_{r=1}^{1}(3r^{5}+r^{3})=3(1)^5+(1)^3=4,\)

and

\(\displaystyle \frac12\cdot 1^3\cdot 2^3=4.\)

So \(H_1\) is true.

Now assume \(H_k\) is true for some \(k\ge 1\), so that

\(\displaystyle \sum_{r=1}^{k}(3r^{5}+r^{3})=\frac12 k^{3}(k+1)^{3}.\)

Then

\(\displaystyle \sum_{r=1}^{k+1}(3r^{5}+r^{3})=\sum_{r=1}^{k}(3r^{5}+r^{3})+3(k+1)^5+(k+1)^3.\)

Using the inductive hypothesis,

\(\displaystyle \sum_{r=1}^{k+1}(3r^{5}+r^{3})=\frac12 k^{3}(k+1)^{3}+3(k+1)^5+(k+1)^3.\)

Factor out \((k+1)^3\):

\(\displaystyle =(k+1)^3\left(\frac12 k^3+3(k+1)^2+1\right).\)

Now simplify the bracket:

\(\displaystyle \frac12 k^3+3(k^2+2k+1)+1=\frac12 k^3+3k^2+6k+4.\)

Also

\(\displaystyle \frac12(k+2)^3=\frac12(k^3+6k^2+12k+8),\)

so

\(\displaystyle \frac12 k^3+3k^2+6k+4=\frac12(k+2)^3.\)

Hence

\(\displaystyle \sum_{r=1}^{k+1}(3r^{5}+r^{3})=\frac12 (k+1)^3(k+2)^3.\)

This is exactly \(H_{k+1}\), so the result follows by induction for all \(n\ge 1\).

Now use the result to find \(\sum_{r=1}^n r^5\). Since

\(\displaystyle \sum_{r=1}^{n}(3r^{5}+r^{3})=3\sum_{r=1}^{n}r^5+\sum_{r=1}^{n}r^3,\)

and the List of Formulae gives

\(\displaystyle \sum_{r=1}^{n}r^3=\left(\frac{n(n+1)}{2}\right)^2=\frac14 n^2(n+1)^2,\)

we have

\(\displaystyle 3\sum_{r=1}^{n}r^5+\frac14 n^2(n+1)^2=\frac12 n^3(n+1)^3.\)

Therefore

\(\displaystyle 3\sum_{r=1}^{n}r^5=\frac12 n^3(n+1)^3-\frac14 n^2(n+1)^2.\)

Factorising,

\(\displaystyle 3\sum_{r=1}^{n}r^5=\frac14 n^2(n+1)^2\bigl(2n(n+1)-1\bigr).\)

So

\(\displaystyle \sum_{r=1}^{n}r^5=\frac{1}{12}n^2(n+1)^2\bigl(2n^2+2n-1\bigr).\)

Thus \(\mathrm{Q}(n)=2n^2+2n-1\).

Let

\(\displaystyle H_n:\sum_{r=1}^{n}(3r^5+r^3)=\frac12 n^3(n+1)^3.\)

Base case. For \(n=1\),

\(\displaystyle \sum_{r=1}^{1}(3r^5+r^3)=3+1=4,\)

while

\(\displaystyle \frac12\cdot 1^3\cdot 2^3=4.\)

So \(H_1\) is true.

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

\(\displaystyle \sum_{r=1}^{k+1}(3r^5+r^3)=\sum_{r=1}^{k}(3r^5+r^3)+3(k+1)^5+(k+1)^3.\)

Using the inductive hypothesis,

\(\displaystyle \sum_{r=1}^{k+1}(3r^5+r^3)=\frac12 k^3(k+1)^3+3(k+1)^5+(k+1)^3.\)

Factor out \((k+1)^3\):

\(\displaystyle = (k+1)^3\left(\frac12 k^3+3(k+1)^2+1\right).\)

Now simplify the bracket:

\(\displaystyle \frac12 k^3+3(k+1)^2+1=\frac12 k^3+3(k^2+2k+1)+1\)

\(\displaystyle =\frac12 k^3+3k^2+6k+4.\)

But

\(\displaystyle \frac12(k+2)^3=\frac12(k^3+6k^2+12k+8)=\frac12 k^3+3k^2+6k+4.\)

Hence

\(\displaystyle \sum_{r=1}^{k+1}(3r^5+r^3)=\frac12 (k+1)^3(k+2)^3,\)

which is exactly \(H_{k+1}\). Therefore, by induction,

\(\displaystyle \sum_{r=1}^{n}(3r^5+r^3)=\frac12 n^3(n+1)^3\)

for all \(n\ge 1\).

Now use the formula \(\displaystyle \sum_{r=1}^{n}r^3=\left(\frac{n(n+1)}{2}\right)^2=\frac14 n^2(n+1)^2\).

Since

\(\displaystyle \sum_{r=1}^{n}(3r^5+r^3)=3\sum_{r=1}^{n}r^5+\sum_{r=1}^{n}r^3,\)

we get

\(\displaystyle 3\sum_{r=1}^{n}r^5+\frac14 n^2(n+1)^2=\frac12 n^3(n+1)^3.\)

So

\(\displaystyle 3\sum_{r=1}^{n}r^5=\frac12 n^3(n+1)^3-\frac14 n^2(n+1)^2\)

\(\displaystyle =\frac14 n^2(n+1)^2\bigl(2n(n+1)-1\bigr).\)

Therefore

\(\displaystyle \sum_{r=1}^{n}r^5=\frac{1}{12}n^2(n+1)^2(2n^2+2n-1).\)

So \(\mathrm{Q}(n)=2n^2+2n-1\).