Question

NO LINKS!! Please help me with this problem. Part 8ff​

Answer 1

`(1)/(36n^2+6n)`

Explanation:

Given factorial expression:

`((6n-1)!)/((6n+1)!)`

`\boxed{\begin{minipage}{6cm}\underline{Factorial Rule}\\\\$n!=\:n\cdot \left(n-1\right) \cdot \left(n-2\right) \cdot ... \cdot 3 \cdot 2\cdot 1$\\ \end{minipage}}`

Apply the factorial rule to the numerator and denominator of the given rational factorial expression:

`(6n-1)!=\left(6n-1\right)\cdot \left(6n-2\right)\cdot \left(6n-3\right)\cdot... \cdot 3 \cdot 2\cdot 1`

`\left(6n+1\right)!=\left(6n+1\right)\cdot \:6n \cdot (6n-1) \cdot...\cdot 3 \cdot 2\cdot 1`

Therefore:

`\begin{aligned}\implies ((6n-1)!)/((6n+1)!)&=(\left(6n-1\right)\cdot \left(6n-2\right)\cdot \left(6n-3\right)\cdot... \cdot 3 \cdot 2\cdot 1)/(\left(6n+1\right)\cdot \:6n \cdot (6n-1) \cdot...\cdot 3 \cdot 2\cdot 1)\\\\&=(1)/((6n+1) \cdot 6n)\\\\&=(1)/(6n(6n+1))\\\\&=(1)/(36n^2+6n)\end{aligned}`

Answer 2

• 
  `\cfrac{1}{6n(6n+1)}`

--------------------------------

We know that:

• n! = 1·2·3·4·...·n

Therefore:

• (6n + 1)! = (6n - 1)!·6n·(6n + 1)

Therefore:

• 
  `\cfrac{(6n-1)!}{(6n+1)!} =\cfrac{(6n-1)!}{(6n-1)!(6n)(6n+1)} =\cfrac{1}{6n(6n+1)}`