Given that $1-2r = A(r+1) + Br$, find the constants $A$ and $B$. Use the method of differences to find $\displaystyle \sum_{r=1}^n \frac{1-2r}{3^r}$. Hence find the value of $\displaystyle \sum_{r=1}^{\infty} \frac{2-2r}{3^r}$.

1. Substitute a suitable value for $r$ to eliminate $B$ and find $A$

Substitute $r = 0$ into the identity $1 - 2r = A(r+1) + Br$:
$1 - 2(0) = A(0+1) + B(0)$
$1 = A$
$A = 1$

2. Substitute another suitable value for $r$ to eliminate $A$ and find $B$

Substitute $r = -1$ into the identity:
$1 - 2(-1) = A(0) + B(-1)$
$1 + 2 = -B$
$3 = -B$
$B = -3$

1. Determine the constants $A$ and $B$ from part (i) to rewrite the general term into a difference of two terms.

From part (i), solving $1-2r = A(r+1) + Br$ yields $A = 1$ and $B = -3$, giving $1-2r = (r+1) - 3r$.

Substituting this into the general term:
$\frac{1-2r}{3^r} = \frac{(r+1) - 3r}{3^r}$
$= \frac{r+1}{3^r} - \frac{3r}{3^r}$
$= \frac{r+1}{3^r} - \frac{r}{3^{r-1}}$

2. List out the first few terms and the last term of the summation to observe the cancellation pattern.

$\sum_{r=1}^n \frac{1-2r}{3^r} = \sum_{r=1}^n \left( \frac{r+1}{3^r} - \frac{r}{3^{r-1}} \right)$
$= \left( \frac{2}{3^1} - \frac{1}{3^0} \right)$
$+ \left( \frac{3}{3^2} - \frac{2}{3^1} \right)$
$+ \left( \frac{4}{3^3} - \frac{3}{3^2} \right)$
$+ \dots$
$+ \left( \frac{n+1}{3^n} - \frac{n}{3^{n-1}} \right)$

3. Cancel out the terms systematically to deduce the final sum.

By observing the cancellation diagonally, all intermediate terms cancel out, leaving only the negative part of the first term and the positive part of the last term:
$= \frac{n+1}{3^n} - \frac{1}{3^0}$
$= \frac{n+1}{3^n} - 1$

1. Concisely establish the results of parts (i) and (ii) required to evaluate the infinite series.

From part (i), substituting $r=0$ and $r=-1$ into $1-2r = A(r+1) + Br$ yields $A=1$ and $B=-3$.
From part (ii), using the method of differences:
$\sum_{r=1}^n \frac{1-2r}{3^r} = \sum_{r=1}^n \frac{(r+1) - 3r}{3^r} = \sum_{r=1}^n \left( \frac{r+1}{3^r} - \frac{r}{3^{r-1}} \right)$
By writing out the first few and last few terms, we observe that consecutive terms cancel out in pairs:
$= \left( \frac{2}{3} - 1 \right) + \left( \frac{3}{3^2} - \frac{2}{3} \right) + \dots + \left( \frac{n+1}{3^n} - \frac{n}{3^{n-1}} \right)$
$= \frac{n+1}{3^n} - 1$

2. Relate the target infinite series to the series evaluated in part (ii).

We can express the numerator of the target sequence as $(1-2r) + 1$, allowing us to algebraically split the sum:
$\frac{2-2r}{3^r} = \frac{1-2r}{3^r} + \frac{1}{3^r}$
Therefore, the required infinite sum can be split into two separate infinite series:
$\sum_{r=1}^{\infty} \frac{2-2r}{3^r} = \sum_{r=1}^{\infty} \frac{1-2r}{3^r} + \sum_{r=1}^{\infty} \left(\frac{1}{3}\right)^r$

3. Evaluate the sum to infinity for the series found in part (ii).

Using the expression obtained for the sum to $n$ terms, we take the limit as $n \to \infty$:
$\sum_{r=1}^{\infty} \frac{1-2r}{3^r} = \lim_{n \to \infty} \left( \frac{n+1}{3^n} - 1 \right)$
Since the exponential denominator $3^n$ grows much faster than the linear numerator $n+1$, $\lim_{n \to \infty} \frac{n+1}{3^n} = 0$.
$\sum_{r=1}^{\infty} \frac{1-2r}{3^r} = 0 - 1 = -1$

4. Evaluate the infinite geometric series and compute the final combined value.

The series $\sum_{r=1}^{\infty} \left(\frac{1}{3}\right)^r$ is an infinite geometric progression with first term $a = \frac{1}{3}$ and common ratio $r = \frac{1}{3}$.
Using the formula for the sum to infinity:
$\sum_{r=1}^{\infty} \left(\frac{1}{3}\right)^r = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$
Adding the two limiting sums together gives the final value:
$\sum_{r=1}^{\infty} \frac{2-2r}{3^r} = -1 + \frac{1}{2} = -\frac{1}{2}$