Answer: (a) \(f(r+1)-f(r)=r(r+1)!-(r-1)r!=r!\bigl(r^2+1\bigr)\).
Hence \(r!\bigl(r^2+1\bigr)=f(r+1)-f(r)\), so the sum telescopes:
\(\displaystyle \sum_{r=n+1}^{2n} r!\bigl(r^2+1\bigr)=\sum_{r=n+1}^{2n}\bigl(f(r+1)-f(r)\bigr)=f(2n+1)-f(n+1)\).
Therefore
\(\displaystyle \sum_{r=n+1}^{2n} r!\bigl(r^2+1\bigr)=2n(2n+1)!-n(n+1)!\).
First simplify \(f(r+1)-f(r)\) using \(f(r)=r!(r-1)\).
We have \(f(r+1)=(r+1)!\cdot r=r(r+1)!\), and \(f(r)=r!(r-1)\). So
\(\begin{aligned} f(r+1)-f(r) &=r(r+1)!-(r-1)r!\\ &=r\cdot(r+1)r!- (r-1)r!\\ &=r!\bigl(r(r+1)-(r-1)\bigr)\\ &=r!\bigl(r^2+r-r+1\bigr)\\ &=r!\bigl(r^2+1\bigr). \end{aligned}\)
So \(r!\bigl(r^2+1\bigr)=f(r+1)-f(r)\), which is ideal for a telescoping sum.
Now
\(\begin{aligned} \sum_{r=n+1}^{2n} r!\bigl(r^2+1\bigr) &=\sum_{r=n+1}^{2n} \bigl(f(r+1)-f(r)\bigr)\\ &=(f(n+2)-f(n+1))+(f(n+3)-f(n+2))+\cdots +(f(2n+1)-f(2n))\\ &=f(2n+1)-f(n+1). \end{aligned}\)
Finally, substitute into \(f(r)=r!(r-1)\):
\(f(2n+1)=(2n+1)!(2n)=2n(2n+1)!\), and \(f(n+1)=(n+1)!\,n=n(n+1)!\).
Hence
\(\displaystyle \sum_{r=n+1}^{2n} r!\bigl(r^2+1\bigr)=2n(2n+1)!-n(n+1)!\).
