A sequence $u_1, u_2, u_3, \dots$ is such that $u_1 = -1$ and $u_{n+1} = u_n + 2(3^{-n}) + a$, where $n \in \mathbb{Z}^+$ and $a$ is a constant. By considering $u_{n+1} - u_n$, find $u_n$ in terms of $n$ and $a$. Prove by induction that, for every positive integer $n$,
$(1^3 + 3 \cdot 1^5) + (2^3 + 3 \cdot 2^5) + (3^3 + 3 \cdot 3^5) + \dots + (n^3 + 3n^5) = \frac{1}{2}n^3(n+1)^3.$ Given that $\sum_{r=1}^{n} r^3 = \frac{n^2}{4}(n+1)^2$, find $\sum_{r=1}^{2n} r^5$ in terms of $n$.

1. Set up the difference equation and apply summation from $r=1$ to $n-1$.

Given $u_{n+1} - u_n = 2(3^{-n}) + a$, we consider the sum of differences from $r=1$ to $n-1$:
$\sum_{r=1}^{n-1} (u_{r+1} - u_r) = \sum_{r=1}^{n-1} \left[ 2(3^{-r}) + a \right]$

2. Evaluate the left-hand side using the method of differences.

Evaluating the telescoping sum on the left-hand side:
$\sum_{r=1}^{n-1} (u_{r+1} - u_r) = (u_2 - u_1) + (u_3 - u_2) + \dots + (u_n - u_{n-1})$
$= u_n - u_1$

3. Evaluate the right-hand side using the sum of a geometric progression.

The right-hand side consists of a geometric series and a constant sum. For the geometric series, the first term is $3^{-1} = \frac{1}{3}$, the common ratio is $r = \frac{1}{3}$, and there are $n-1$ terms:
$\sum_{r=1}^{n-1} \left[ 2(3^{-r}) + a \right] = 2 \sum_{r=1}^{n-1}(3^{-r}) + \sum_{r=1}^{n-1}a$
$= 2 \left[ \frac{\frac{1}{3} \left( 1 - \left(\frac{1}{3}\right)^{n-1} \right)}{1 - \frac{1}{3}} \right] + a(n-1)$
$= 2 \left[ \frac{\frac{1}{3} (1 - 3^{-(n-1)})}{\frac{2}{3}} \right] + a(n-1)$
$= 1 - 3^{1-n} + a(n-1)$

4. Equate the evaluated sides and substitute the given $u_1$ to find $u_n$.

Equating the two sides and substituting $u_1 = -1$:
$u_n - (-1) = 1 - 3^{1-n} + a(n-1)$
$u_n + 1 = 1 - 3^{1-n} + a(n-1)$
$u_n = a(n-1) - 3^{1-n}$

1. Show that the formula holds for the base case $n=1$.

Let $P_n$ be the proposition that $\sum_{r=1}^n (r^3 + 3r^5) = \frac{1}{2}n^3(n+1)^3$.
When $n=1$,
$\text{LHS} = 1^3 + 3(1^5) = 4$
$\text{RHS} = \frac{1}{2}(1)^3(1+1)^3 = \frac{1}{2}(8) = 4$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. Assume the formula holds for $n=k$ and set up the sum for $n=k+1$.

Assume $P_k$ is true for some positive integer $k$. That is,
$\sum_{r=1}^k (r^3 + 3r^5) = \frac{1}{2}k^3(k+1)^3$
For $n = k+1$, we consider the sum of the first $k+1$ terms:
$\sum_{r=1}^{k+1} (r^3 + 3r^5) = \sum_{r=1}^k (r^3 + 3r^5) + (k+1)^3 + 3(k+1)^5$
$= \frac{1}{2}k^3(k+1)^3 + (k+1)^3 + 3(k+1)^5$

3. Factorise and simplify the expression to show it matches the formula for $n=k+1$.

Factorise $\frac{1}{2}(k+1)^3$ from the expression:
$= \frac{1}{2}(k+1)^3 \left[ k^3 + 2 + 6(k+1)^2 \right]$
Expand the quadratic term inside the bracket:
$= \frac{1}{2}(k+1)^3 \left[ k^3 + 2 + 6(k^2 + 2k + 1) \right]$
$= \frac{1}{2}(k+1)^3 ( k^3 + 6k^2 + 12k + 8 )$
Recognise the cubic expansion of $(k+2)^3$:
$= \frac{1}{2}(k+1)^3 (k+2)^3$
$= \frac{1}{2}(k+1)^3 \left((k+1)+1\right)^3$
This is exactly the formula for $P_{k+1}$, so $P_{k+1}$ is true if $P_k$ is true.

4. State the formal conclusion of the mathematical induction.

Since $P_1$ is true, and $P_k$ is true implies $P_{k+1}$ is true, by Mathematical Induction, $P_n$ is true for all positive integers $n$.

1. Relate the summation of $r^5$ to the summation from part (b) and the given sum of $r^3$.

From part (b), we established that:
$\sum_{r=1}^{n} (r^3 + 3r^5) = \frac{1}{2}n^3(n+1)^3$
Using the property of summations, we can split this into two separate series:
$\sum_{r=1}^{n} r^3 + 3 \sum_{r=1}^{n} r^5 = \frac{1}{2}n^3(n+1)^3$

2. Substitute the given formula for the sum of cubes and solve for the sum of $r^5$.

We are given $\sum_{r=1}^{n} r^3 = \frac{n^2}{4}(n+1)^2$. Substituting this into the equation:
$\frac{n^2}{4}(n+1)^2 + 3 \sum_{r=1}^{n} r^5 = \frac{1}{2}n^3(n+1)^3$
$3 \sum_{r=1}^{n} r^5 = \frac{1}{2}n^3(n+1)^3 - \frac{n^2}{4}(n+1)^2$
Factorising by taking out the common factor $\frac{n^2}{4}(n+1)^2$:
$3 \sum_{r=1}^{n} r^5 = \frac{n^2}{4}(n+1)^2 [2n(n+1) - 1]$
$3 \sum_{r=1}^{n} r^5 = \frac{n^2}{4}(n+1)^2 (2n^2 + 2n - 1)$
$\sum_{r=1}^{n} r^5 = \frac{n^2}{12}(n+1)^2 (2n^2 + 2n - 1)$

3. Substitute $2n$ into the general formula to find the sum up to $2n$ terms.

To find $\sum_{r=1}^{2n} r^5$, we replace every instance of $n$ in our formula with $2n$:
$\sum_{r=1}^{2n} r^5 = \frac{(2n)^2}{12}(2n+1)^2 [2(2n)^2 + 2(2n) - 1]$

4. Simplify the expression to obtain the final answer.

Expanding the inner terms and simplifying the fraction:
$\sum_{r=1}^{2n} r^5 = \frac{4n^2}{12}(2n+1)^2 (8n^2 + 4n - 1)$
$\sum_{r=1}^{2n} r^5 = \frac{1}{3}n^2(2n+1)^2(8n^2 + 4n - 1)$