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Question
Simplify `1/("n"!) - 3/(("n" + 1)!) - ("n"^2 - 4)/(("n" + 2)!)`
Solution
`1/("n"!) - 3/(("n" + 1)!) - ("n"^2 - 4)/(("n" + 2)!)`
= `1/("n"!) - 3/(("n" + 1)"n"!) - (("n" - 2)("n" + 2))/(("n" + 2)("n" + 1)"n"!)`
= `1/("n"!)[1-3/("n" + 1)-("n" - 2)/("n" + 1)]`
= `1/("n"!)[("n" + 1 -3 - "n" + 2)/("n" + 1)]`
= `1/("n"!)xx0`
= 0
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