A difference equation and a sum

First, we'll find the solution to the difference equation, which can be written as

$y_{k+1} - \dfrac{3}{4} y_k = \dfrac{1}{4}$

This is a simple first order difference equation whose solution is known to be

$y_k = A \left( \dfrac{3}{4} \right) ^k + B$

where the constants $A$ , $B$ are yet to be determined.

Plugging in this form into the difference equation yields,

$A \left( \dfrac{3}{4} \right) ^{k+1} + B - \dfrac{3}{4} \left( A \left( \dfrac{3}{4} \right) ^k + B \right) = \dfrac{1}{4}$

From which, $B = 1$

To find $A$ we use the initial conditions.

$y_2 = \dfrac{1}{4} = A \left( \dfrac{3}{4} \right) ^2 + 1$

from which $A = - \dfrac{4}{3}$

Therefore, $y_k = 1 - \left( \dfrac{3}{4} \right) ^{k-1}$

Next, we'll find $x_k$ . Substituting $y_k$ into $x_{k+1} = \dfrac{1}{4} ( 1 - y_k )$ gives us

$x_{k+1} = \dfrac{1}{4} \left( \dfrac{3}{4} \right) ^{k-1}$

so that,

$x_{k} = \dfrac{1}{4} \left( \dfrac{3}{4} \right) ^{k-2}$

Now, we want to find the sum,

$S = \displaystyle \sum_{k=2}^{\infty} k x_k = \dfrac{1}{4} \sum_{k=2}^{\infty} k \left( \dfrac{3}{4} \right) ^{k-2} =\dfrac{1}{4} \sum_{j=0}^{\infty} (j+2) \left( \dfrac{3}{4} \right) ^{j}$

Simplifying,

$S =\displaystyle \dfrac{1}{4} \left( 2 \sum_{j=0}^{\infty}\left( \dfrac{3}{4} \right) ^{j} + \sum_{j=0}^{\infty} j \left( \dfrac{3}{4} \right) ^{j} \right)$

The first sum is just a geometric series sum and evaluates to $\dfrac{1}{1 - \frac{3}{4} } = 4$ . The second sum is handled as follows

$\displaystyle \sum_{j=0}^{\infty} j \left( \dfrac{3}{4} \right) ^{j} = \dfrac{3}{4} \sum_{j=1}^{\infty} j \left( \dfrac{3}{4} \right) ^{j-1}$ $\displaystyle =\dfrac{3}{4} \dfrac{d}{ds} \left( \sum_{j=1}^{\infty} s^j \right) \bigg|_{s = \frac{3}{4} } = \dfrac{3}{4} \dfrac{d}{ds} \left( \dfrac{s}{1 - s} \right) \bigg|_{s = \frac{3}{4}} = \dfrac{3}{4} \left( \dfrac{1}{(1 - s)^2} \right) \bigg|_{s = \frac{3}{4} } = \dfrac{3}{4} \times 16 = 12$

Putting it all together,

$S = \dfrac{1}{4} ( 2 \times 4 + 12 ) = \boxed{5}$