Question

If 1(0!)+3.(1!)+7(2!)+13(3!)+21(4!)+⋯ n terms = (4000)(4000!) Then what is the value of n.

Answer 1

$n = 4000$

Explanation:

$Given, 1(0!)+3.(1!)+7(2!)+13(3!)+21(4!)+⋯ n \: \: terms \\ we \: \: see \: \: that \\t_{n} = ( {n}^{2} - n + 1 )(n - )! \\ NOW, \: \: \sum \: t_{n} = \sum \: ( {n}^{2} - n + 1 ) \: (n - 1)! \\ = \sum \: \ \{ {n}^{2} (n - 1)! - (n - 1)(n - 1)!\} \\ \sum \: \ \{ n.n! - (n - 1)(n - 1)!\} \\ = n.n! \\ again \: \: \: \: given \\ 1(0!)+3.(1!)+7(2!)+13(3!)+21(4!)+⋯ n \: \: terms \\ = 4000(4000!) \\ so \: \: \: n = 4000.$