Show that if n is a positive integer with n ≥ 3, then
C(n, n − 2) = ((3n − 1)C(n, 3))/4
"=2\\dfrac{n!}{3!(n-3)!}+3\\dfrac{n!}{4!(n-4)!}"
"=\\dfrac{n(n-1)(n-2)}{3}+\\dfrac{n(n-1)(n-2)(n-3)}{8}"
"=\\dfrac{n^3}{3}-n^2+\\dfrac{2n}{3}+\\dfrac{n^4}{8}-\\dfrac{3n^3}{4}+\\dfrac{11n^2}{8}-\\dfrac{3n}{4}"
"=\\dfrac{n^4}{8}-\\dfrac{5n^3}{12}+\\dfrac{3n^2}{8}-\\dfrac{n}{12}"
"\\dfrac{(3n-1)C(n,3)}{4}=\\dfrac{(3n-1)n!}{3!(n-3)!(4)}"
"=\\dfrac{(3n-1)(n)(n-1)(n-2)}{24}"
"=\\dfrac{n^4}{8}-\\dfrac{5n^3}{12}+\\dfrac{3n^2}{8}-\\dfrac{n}{12}"
Hence
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