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((n+2)!-(n+1)!)/(n!)=...

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

A

`(n+2)!`

B

`(n+1)!`

C

`(n+2)^(2)`

D

`(n+1)^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
Expand, cance, and simplify :
`((n+2)!-(n+1)!)/(n!)=((n+2)!)/(n!)-((n+1)!)/(n!)`
`= (n+2)(n+1)-(n+1)`
`= n^(2)+3n + 2- n - 1`
`= n^(2)+2n + 1`
That's the same as (D), `(n+1)^(2)`.
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