If $\alpha =$minimum of $\left(x^2+2x+3\right)$ and $\beta =\underset{n\to \infty }{Lt}{\displaystyle \sum _{r=1}^n\frac{1}{\left(r+2\right)r!}}$, then ${\displaystyle \sum _{r=0}^n{\alpha }^r{\beta }^{n-2}=}$

$\alpha =$minimum of $x^2+2x+3=\frac{\mathrm{4.1.3}-2^2}{4}=2$

$\beta =\underset{n\to \infty }{Lt}{\displaystyle \sum _{r=1}^n\frac{1}{\left(r+2\right)r!}}$

$=\underset{n\to \infty }{Lt}{\displaystyle \sum _{r=1}^n\frac{r+1}{\left(r+2\right)\left(r+1\right)r!}}=\underset{n\to \infty }{Lt}{\displaystyle \sum _{r=1}^n\frac{\left(r+2\right)-1}{\left(r+2\right)!}}$

$=\underset{n\to \infty }{Lt}{\displaystyle \sum _{r=1}^n\left(\frac{1}{\left(r+1\right)!}-\frac{1}{\left(r+2\right)!}\right)}$

$=\underset{n\to \infty }{Lt}\left[\frac{1}{2}-\frac{1}{\left(n+2\right)!}\right]=\frac{1}{2}$

Now ${\displaystyle \sum _{r=0}^n{\alpha }^r{\beta }^{n-r}}$

$={\beta }^n+\alpha {\beta }^{n-1}+{\alpha }^2{\beta }^{n-2}+\mathrm{....}+{\alpha }^n$

$={\beta }^n\left[1+\frac{\alpha }{\beta }+{\left(\frac{\alpha }{\beta }\right)}^2+\mathrm{.....}+{\left(\frac{\alpha }{\beta }\right)}^n\right]$

$={\beta }^n\frac{{\left(\frac{\alpha }{\beta }\right)}^{n+1}-1}{\frac{\alpha }{\beta }-1}=\frac{1}{2^n}\left(\frac{4^{n+1}-1}{4-1}\right)=\frac{4^{n+1}-1}{{3.2}^n}$