Sumtegrals! - 2

Through partial fraction decomposition and integration, we get

$\displaystyle K=\sum_{n=0}^\infty\frac{9^n}{n!}\left(\frac{1}{n+2}-\frac{1}{n+3}\right)=\sum_{n=0}^\infty\frac{9^n}{n!(n+2)}-\sum_{n=0}^\infty\frac{9^n}{n!(n+3)}$

$\displaystyle=\frac{d}{dx}\left(\sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+2)!}\right)\bigg|_{x=9}-\frac{d^2}{dx^2}\left(\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+3)!}\right)\bigg|_{x=9}$

$\displaystyle=\frac{d}{dx}\left(\frac{1}{x}\sum_{n=0}^{\infty}\frac{x^n}{n!}-\frac{1}{x}-1\right)\bigg|_{x=9}-\frac{d^2}{dx^2}\left(\frac{1}{x}\sum_{n=0}^\infty \frac{x^n}{n!}+\frac{1}{x}+1+\frac{x}{2}\right)\bigg|_{x=9}$

$\displaystyle=\frac{d}{dx}\left(\frac{e^x}{x}-\frac{1}{x}-1\right)\bigg|_{x=9}-\frac{d^2}{dx^2}\left(\frac{e^x}{x}-\frac{1}{x}-1-\frac{x}{2}\right)\bigg|_{x=9}$

$\displaystyle=\left(\frac{e^x}{x}-\frac{1}{x^2}+\frac{1}{x^2}\right)\bigg|_{x=9}-\left(\frac{e^x}{x}-\frac{2e^x}{x^2}+\frac{2e^x}{x^3}-\frac{2}{x^3}\right)\bigg|_{x=9}$

$\displaystyle=\frac{1}{81}(1+8e^9)+\frac{1}{729}(2-65e^9)=\frac{1}{729}(7e^9+11).$

Therefore, the answer is $729+9+7+11+1=\boxed{757}$ .