证明:2Cn2+9Cn3+12Cn4+5Cn5=24分之n^(n+2)(n+1)(n-1)附:组合题,请懂得的回答哈,鄙人不才,

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证明:2Cn2+9Cn3+12Cn4+5Cn5=24分之n^(n+2)(n+1)(n-1)附:组合题,请懂得的回答哈,鄙人不才,
证明:2Cn2+9Cn3+12Cn4+5Cn5=24分之n^(n+2)(n+1)(n-1)
附:组合题,请懂得的回答哈,鄙人不才,

证明:2Cn2+9Cn3+12Cn4+5Cn5=24分之n^(n+2)(n+1)(n-1)附:组合题,请懂得的回答哈,鄙人不才,
2C[n,2]+9C[n,3]+12C[n,4]+5C[n,5]
=2n(n-1)/2!+9n(n-1)(n-2)/3!+12n(n-1)(n-2)(n-3)/4!++5n(n-1)(n-2)(n-3)(n-4)(n-5)/5!
=n(n-1)+3n(n-1)(n-2)/2+n(n-1)(n-2)(n-3)/2+n(n-1)(n-2)(n-3)(n-4)/24
={24n(n-1)+36n(n-1)(n-2)+12n(n-1)(n-2)(n-3)+n(n-1)(n-2)(n-3)(n-4)}/24
整理得:
=(n^5+2n^4-n^3-2n^2)/24
=n^2(n-1)(n+1)(n+2)/24