Write the following in terms of factorials. A) n(n+1)(n+2)(n+3) B) (n+6)(n+7)(n+8) c)n*(n-1)!

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a) You should multiply and divide the given product by factorial `(n-1)!`  such that:


You need to remember that `(n+3)! = 1*2*3*...*(n-1)*n*(n+1)(n+2)(n+3) = (n+2)!(n+3) = (n+1)!*(n+2)*(n+3) = n!*(n+1)*(n+2)*(n+3) = (n-1)!*n*(n+1)*(n+2)*(n+3) `

Hence, you may substitute `(n+3)! for ((n-1)!*n*(n+1)*(n+2)*(n+3))`  such that:

`n*(n+1)*(n+2)*(n+3) = ((n+3)!)/((n-1)!) `

Hence, using factorials, you may write the product `n*(n+1)*(n+2)*(n+3)`  such that: `n*(n+1)*(n+2)*(n+3) = ((n+3)!)/((n-1)!) `

b) Reasoning by analogy yields:

`(n+6)(n+7)(n+8) = ((n+5)!(n+6)(n+7)(n+8))/((n+5)!)`

`(n+6)(n+7)(n+8) = ((n+8)!)/((n+5)!)`

Hence, using factorials, you may write the product `(n+6)(n+7)(n+8)`  such that: `(n+6)(n+7)(n+8) = ((n+8)!)/((n+5)!).`

c) You need to arrange the terms of product such that:

`(n-1)!*n = n!`

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