Q5. Evaluate \frac{(4n+1)!}{(4n-1)!}

To simplify the expression, we can cancel out like terms in the numerator and denominator:

\frac{(4n+1)!}{(4n-1)!} = \frac{(4n+1)(4n)(4n-1)!}{(4n-1)!}

Here, the (4n-1)! terms in the numerator and denominator cancel out:

= (4n+1)(4n)

= 16n^2 + 4n

Therefore, \frac{(4n+1)!}{(4n-1)!} simplifies to 16n^2 + 4n.