I Was Very Amazed At The Solution 14

It is difficult of course to list all 50 terms and add them one by one. So, first, find the characteristic polynomial of the recurrence equation. We have:

$\\ 3r^2 = 2r + 1 \implies r \in \left\{ 1, -\dfrac{1}{3} \right\}$

Hence, $x_k = (\alpha _1)(1^k) + (\alpha _2)\left(-\dfrac{1}{3}\right)^k$ .

To solve for the constants $\alpha_1$ and $\alpha_2$ , we need to substitute the given.

For $k = 1 \implies x_1 = 1 = \alpha_1 - \dfrac{\alpha_2}{3}$ .

For $k = 2 \implies x_2 = 6 = \alpha_1 + \dfrac{\alpha_2}{9}$ .

subtracting the 2 equations yields: $5 = (\alpha_2)\left(\dfrac{1}{9} + \dfrac{1}{3}\right) \implies \alpha_2 = 5\left(\dfrac{9}{4}\right) = \dfrac{45}{4}$ .

Then, the value of $\alpha_1 = 1 + \dfrac{\left(\dfrac{45}{4}\right)}{3} = 1 + \dfrac{15}{4} = \dfrac{19}{4}$ .

Therefore, $\boxed{x_k = \left(\dfrac{19}{4}\right)(1^k) + \left(\dfrac{45}{4}\right)\left(-\dfrac{1}{3}\right)^k}$ .

We found already the general term for $x_k$ . hence,

$\displaystyle \begin{aligned} \sum_{k = 1}^{50} x_k &= \sum_{k = 1}^{50} \left( \left(\dfrac{19}{4}\right)(1^k) + \left(\dfrac{45}{4}\right)\left(-\dfrac{1}{3}\right)^k \right) \\ &= \sum_{k = 1}^{50} \left( \left(\dfrac{19}{4}\right)(1^k) \right) + \sum_{k = 1}^{50} \left( \left(\dfrac{45}{4}\right)\left(-\dfrac{1}{3}\right)^k \right) \\ &= \left(\dfrac{19}{4}\right)(50) + \left(\dfrac{45}{4}\right) \sum_{k = 1}^{50} \left( \left(-\dfrac{1}{3}\right)^k \right) \\ & = \dfrac{475}{2} + \left(\dfrac{45}{4}\right) \left(\dfrac{\left(-\dfrac{1}{3}\right)\left( 1-\left(-\dfrac{1}{3}\right)^{50}\right) }{1- \left( -\dfrac{1}{3}\right)} \right) \\ &= \dfrac{475}{2} - \left(\dfrac{45}{4}\right) \left(\dfrac{\left( 1-\dfrac{1}{3^{50}}\right)}{4} \right) \\ & = \dfrac{475}{2} - \left(\dfrac{45}{4}\right) \left(\dfrac{3^{50} - 1}{(3^{50})(4)} \right) \approx \dfrac{475}{2} - \left(\dfrac{45}{4}\right) \left(\dfrac{3^{50}}{(3^{50})(4)} \right) \\ &\approx \dfrac{475}{2} - \left(\dfrac{45}{4}\right) \left(\dfrac{1}{4} \right) \approx \dfrac{475}{2} - \dfrac{45}{16} \\ & \approx \boxed{234.6875} . \end{aligned}$