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· · 个人记录

\begin{aligned} F(x)&=\sum\limits_{i\ge 0}f_ix^i\\ &=1+\sum\limits_{i\ge 1}\left(f_{i-1}+(i-1)+\binom{i-1}{3}\right)x^i\\ &=1+x\sum\limits_{i\ge 1}\left(f_{i-1}x^{i-1}+(i-1)x^{i-1}+\binom{i-1}{3}x^{i-1}\right)\\ &=1+x\left(\sum\limits_{i\ge 1}f_{i-1}x^{i-1}+\sum\limits_{i\ge 1}(i-1)x^{i-1}+\sum\limits_{i\ge 1}\binom{i-1}{3}x^{i-1}\right)\\ &=1+x\left(\sum\limits_{i\ge 0}f_{i}x^{i}+\sum\limits_{i\ge 0}ix^i+\sum\limits_{i\ge 0}\binom{i}{3}x^i\right)\\ &=1+x\left(F(x)+\sum\limits_{i\ge 0}ix^i+\sum\limits_{i\ge 0}\dfrac{i(i-1)(i-2)}{6}x^i\right)\\ &=1+x\left(F(x)+\dfrac{1}{(1-x)^2}+\dfrac{x^3}{(1-x)^4}\right)\\ &=1+xF(x)+\dfrac{x}{(1-x)^2}+\dfrac{x^4}{(1-x)^4}\\ &=1+xF(x)+\dfrac{x}{(1-x)^2}+\dfrac{x^4}{(1-x)^4}\\ (1-x)F(x)&=1+\dfrac{x}{(1-x)^2}+\dfrac{x^4}{(1-x)^4}\\ (1-x)F(x)&=\dfrac{(1-x)^4+x(1-x)^2+x^4}{(1-x)^4}\\ F(x)&=\dfrac{(1-x)^4+x(1-x)^2+x^4}{(1-x)^5}\\ &=\left((1-x)^4+x(1-x)^2+x^4\right)(1-x)^{-5}\\ &=\left((1-x)^4+x(1-x)^2+x^4\right)\sum\limits_{i\ge 0}\dfrac{(-5)^{\underline i}}{i!}x^i\\ &=\left((1-x)^4+x(1-x)^2+x^4\right)\sum\limits_{i\ge 0}\binom{i+4}{i}x^i\\ \end{aligned}