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Continuing excercise : Find solutions of recursive equations

$$ x_{n} = 14x_{n-1} - 49x_{n-2} + (n-2)7^{n-2}, n\ge 2\\ a_{n} = 14a_{n-1} - 49a_{n-2} + (n-2)7^{n-2}\\ F(x) = \sum_{n=0}a_nx^n = 1 + \sum_{n=1}a_nx^n\\ F(x) - 1 = \sum_{n=1}a_nx^n$$ \begin{align} F(x) &= 1 + 14x + \sum_{n=2}^\infty (14a_{n-1} - 49a_{n-2} + (n-2)7^{n-2})x^n\\ &= 1 + 14x + 14\sum_{n=2}^\infty a_{n-1}x^n -49\sum_{n=2}^\infty a_{n-2}x^n + \sum_{n=2}^\infty (n-2)7^{n-2}x^n \\ &= 1 + 14x + 14x\sum_{n=2}^\infty a_{n-1}x^{n-1} -49x^2\sum_{n=2}^\infty a_{n-2}x^{n-2} + x^2\sum_{n=2}^\infty (n-2)7^{n-2}x^{n-2} \\ &= 1 + 14x + 14x\sum_{n=1}^\infty a_{n}x^{n} -49x^2\sum_{n=0}^\infty a_{n}x^{n} + x^27x\sum_{n=0}^\infty n7x^{n-1} \\ &= 1 + 14x + 14x(F(x) - 1) - 49x^2F(x) + \frac{7x^3}{(1-7x)^2}\\ \end{align} $$ F(x)= 1+ 14xF(x) - 49x^2F(x) + \frac{7x^3}{(1-7x)^2} \\ F(x)(1-14x+49x^2) = 1 + \frac{7x^3}{(1-7x)^2} \\ F(x)(1-14x+49x^2) = \frac{7x^3 + (1-7x)^2}{(1-7x)^2} \\ F(x) = \frac{7x^3 + 49x^2 - 14x + 1}{(1-7x)^4} \\ \frac{7x^3 + 49x^2 - 14x + 1}{(1-7x)^4} = \frac{A}{1-7x} + \frac{B}{(1-7x)^2} + \frac{C}{(1-7x)^3} + \frac{D}{(1-7x)^4}\\ 7x^3 + 49x^2 - 14x + 1 = A(1-7x)^3 + B(1-7x)^2 + C(1-7x) + D\\ 7x^3 + 49x^2 - 14x + 1 = A + B + C + D - 343Ax^3 + (147A+49B)x^2 - (21A+14B+7C)x\\ \begin{cases} 7 = -343A \\ 49 = 147A + 49B \\ 14 = 21A + 14B + 7C \\ 1 = A+B+C+D \end{cases}\\ \begin{cases} A = \frac{-1}{49} \\ B = \frac{52}{49} \\ C = \frac{-3}{49} \\ D = \frac{1}{49} \end{cases}\\ F(x) = \frac{\frac{-1}{49}}{1-7x} + \frac{\frac{52}{49}}{(1-7x)^2} + \frac{\frac{-3}{49}}{(1-7x)^3} +\frac{\frac{1}{49}}{(1-7x)^4} $$

But what now? Don't know how transform it back into sum and get $a_n$ as solution.

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    $\begingroup$ Use the fact that $$\frac1{(1-x)^{k+1}}=\sum_n\binom{n+k}nx^n\;,$$ which is established by repeated differentiation of the case $k=0$. $\endgroup$ – Brian M. Scott May 19 at 22:15

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