Question

Define $a_{k}=\left(k^{2} + 1\right)k!$ and $b_{k}=a_{1}+a_{2}+a_{3}+.\ldots \ldots ...+a_{k}.$ Let $\frac{a_{100}}{b_{100}}=\frac{m}{n}$ where $m$ and $n$ are relatively prime numbers. Then the value of $\left(n - m\right)$ is:

Answer 1

$a_k=\left(k^2+1\right) k !=(k(k+1)-(k-1)) k !=k(k+1) !-(k-1) k !$
$a_{1}=1\cdot 2!-0$
$a_{2}=2\cdot 3!-1\cdot 2!$
$a_{3}=3\cdot 4!-2\cdot 3!$
$\vdots$
$\frac{a_{k} = k \left(k + 1\right) ! - \left(k - 1\right) k !}{H e n c e \, b_{k} = k \left(k + 1\right) !}$
$\therefore \frac{a_{k}}{b_{k}}=\frac{\left(k^{2} + 1\right) k !}{k \left(k + 1\right) !}=\frac{\left(k^{2} + 1\right)}{k \left(k + 1\right)}=\frac{k^{2} + 1}{k^{2} + k}$
$\frac{a_{100}}{b_{100}}=\frac{10001}{10100}=\frac{m}{n};\left(n - m\right)=99 \, Ans.$