Let \ a _ n \ _ n = 0 ^ ∞ be a sequence such that a _ 0 = a _ 1 = 0...

Given ${a}_{n + 2} = 2{a}_{n + 1} - {a}_{n} + 1$

Dividing by ${7}^{n + 2}$, we get

$\dfrac{{a}_{n + 2}}{{7}^{n + 2}} = \dfrac{2}{7} \cdot \dfrac{{a}_{n + 1}}{{7}^{n + 1}} - \dfrac{1}{49} \cdot \dfrac{{a}_{n}}{{7}^{n}} + \dfrac{1}{{7}^{n + 2}}$

So, $\mathop{∑}_{n = 2}^{\infty}\dfrac{{a}_{n + 2}}{{7}^{n + 2}} = \dfrac{2}{7}\mathop{∑}_{n = 2}^{\infty}\dfrac{{a}_{n + 1}}{{7}^{n + 1}} - \dfrac{1}{49}\mathop{∑}_{n = 2}^{\infty}\dfrac{{a}_{n}}{{7}^{n}} + \mathop{∑}_{n = 2}^{\infty}\dfrac{1}{{7}^{n + 2}}$

Let
$\mathop{∑}_{n = 2}^{\infty}\dfrac{{a}_{n}}{{7}^{n}} = P$

$⇒ \left(P - \dfrac{{a}_{3}}{{7}^{3}} - \dfrac{{a}_{2}}{{7}^{2}}\right) = \dfrac{2}{7}\left(P - \dfrac{{a}_{2}}{{7}^{2}}\right) - \dfrac{1}{49}P + \dfrac{1}{{7}^{4}}\left(\dfrac{1}{1 - \dfrac{1}{7}}\right)$

\left(∵ {a}_{2} = 1 \text{ and } {a}_{3} = 3\right)$</p><p>$⇒ P\left(1 - \dfrac{2}{7} + \dfrac{1}{49}\right) = \dfrac{3}{{7}^{3}} + \dfrac{1}{{7}^{2}} - \dfrac{2}{{7}^{3}} + \dfrac{1}{6 \cdot {7}^{3}}$</p><p>$⇒ \dfrac{36P}{49} = \dfrac{1}{42} ⇒ P = \dfrac{7}{216}$