Yet another question 4 - Corrected

Note that $a_k$ can be written as :- $a_k=\,(k^{2}+1)\cdot k!\,=\,\left((k+1)^{2}-2k\right)\cdot k!\,=\,(k+1) \cdot (k+1)! - 2k \cdot k!\,=\,\left((k+1)\cdot (k+1)!-k \cdot k!\right)-k\cdot k!$ . Now, $b_n\,=\,\sum_{k=1}^{n}a_k\,=\, \sum_{k=1}^{n} \left( (k+1) \cdot (k+1)! - k \cdot k!\right) - k \cdot k! \,\\ =\,\sum_{k=1}^{n} (k+1) \cdot (k+1)! - k\cdot k!\,-\,\sum_{k=1}^{n} k\cdot k!\,\\ =\, \underbrace{\sum_{k=1}^{n} (k+1) \cdot (k+1)! - k\cdot k! }_{\text{Telescopic Series}}\,-\,\underbrace{\sum_{k=1}^{n} (k+1)! - k!}_{\text{Telescopic Series}}\,\\ =\, \left((n+1) \cdot (n+1)!-1)- \left((n+1)! - 1\right)\right)\,\\ =\,n \cdot (n+1)!$ . Hence, $b_n\,=\,n \cdot (n+1)!$ .

Now, $\large\frac{a_{100}}{b_{100}}\,=\,\frac{(100^{2}+1) \cdot 100!}{100 \cdot 101!}\,=\,\frac{100^{2}+1}{100 \cdot 101}$ .

Hence, $b-a\,=\,100 \cdot 101 - ( 100^{2} + 1 )\,=\,100 \cdot 101 - 100^{2} - 1\,=\, 100(101 - 100) -1 \,=\, 100 -1\,=\, \boxed{99}$ .