Express [latex]\frac{{{r}^{2}}+7r+11}{\left( r+4 \right)!}[/latex] in the form [latex]\frac{A}{\left( r+2 \right)!}+\frac{B}{\left( r+4 \right)!}[/latex] where [latex]A[/latex] and [latex]B[/latex] are constants to be found.
Hence show that [latex]\sum\limits_{r=1}^{n}{\frac{{{r}^{2}}+7r+11}{\left( r+4 \right)!}=\frac{5}{24}-\frac{n+5}{\left( n+4 \right)!}}[/latex].