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gizmo_the_mogwai [7]
1 year ago
11

D(1) = 1,d(n) = n.d(n-1) for n ≥ 2

Mathematics
1 answer:
SIZIF [17.4K]1 year ago
3 0

Assuming you're solving for d(n), we have by substitution

d(n) = n d(n-1) \implies d(n-1) = (n-1) d(n-2) \\\\ ~~~~ \implies d(n) = n(n-1)d(n-2)

d(n) = n d(n-1) \implies d(n-2) = (n-2) d(n-3) \\\\ ~~~~ \implies d(n) = n(n-1)(n-2)d(n-3)

d(n) = n d(n-1) \implies d(n-3) = (n-3) d(n-4) \\\\ ~~~~ \implies d(n) = n(n-1)(n-2)(n-3) d(n-4)

and so on. After k iterations of this, we have the pattern

d(n) = n(n-1)(n-2)\cdots(n-(k-3))(n-(k-2))(n-(k-1)) d(n-k)

so that after k=n-1 iterations,

d(n) = n(n-1)(n-2)\cdots(n-(n-3))(n-(n-2)) d(n-(n-1)) \\\\ ~~~~= n(n-1)(n-2)\cdots3\cdot2\,d(1) \\\\ ~~~~ = \boxed{n!}

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