Simplify the following expressions.
1. n!+(n+1)!+(n+2)!/(n-1)!
2. (2n+1)!(2n-1)!/[(2n)!]^2
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2009-10-08 5:50 am
回答 (2)
2009-10-08 6:18 am
✔ 最佳答案
1. [n!+(n+1)!+(n+2)!]/(n-1)! I guess there is a bracket to contain the first 3 terms, right?= n + n(n + 1) + n(n + 1)(n + 2)
= n(1 + n + 1 + n^2 + 3n + 2)
= n(n^2 + 4n + 4)
= n(n + 2)^2
2. (2n+1)!(2n-1)!/[(2n)!]^2
= [(2n + 1)!/(2n!)][(2n - 1)!/(2n!)]
= (2n + 1)/2n
2009-10-08 6:37 am
1. [n(n-1)!+(n+1)n(n-1)!+(n+2)(n+1)n(n-1)!]/(n-1)!
= n+(n+1)n+(n+2)(n+1)n
= n[1+n+1+(n+2)(n+1)]
= n[(n+2)(1+n+1)]
= n(n+2)^2
2. (2n+1)!(2n-1)!/[(2n)!]^2
= (2n+1)(2n)!(2n-1)!/[(2n)!]^2
= (2n+1)(2n-1)!/(2n)!
= (2n)!(2n-1)!/(2n)!
= (2n-1)!
= n+(n+1)n+(n+2)(n+1)n
= n[1+n+1+(n+2)(n+1)]
= n[(n+2)(1+n+1)]
= n(n+2)^2
2. (2n+1)!(2n-1)!/[(2n)!]^2
= (2n+1)(2n)!(2n-1)!/[(2n)!]^2
= (2n+1)(2n-1)!/(2n)!
= (2n)!(2n-1)!/(2n)!
= (2n-1)!
參考: 我自己, 有錯請指教!!
收錄日期: 2021-04-23 23:18:45
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