Question

lf the nth term of a sequence be tn =(n-2) (n-3)them show that the first three terms of the sequence are zero but the rest of the terms are positive

Answer 1

$The \: \: answer \: \: is \: \: given \: \: below \\ \\ I \: \: think \: \: that \: \: the \: \: n \: th \: \: term \\ should \: \: be \\ \\ t_{n} = (n - 1)(n - 2)(n - 3) \\ \\ Now, \: \: the \: \: first \: \: term, \\ t_{1} = (1 - 1)(1 - 2)( 1 - 3) = 0, \\ \\ the \: \: second \: \: term, \\ t _{2} = (2 - 1)(2 - 2)(2 - 3) = 0 \\ \\ and \\ \\ the \: \: third \: \: term, \\ t _{3} = (3 - 1)( 3 - 2)(3 - 3) = 0 \\ \\ Now, \: \: the \: \: fourth \: \: term, \\ t_{4} = (4 - 1)(4 - 2)(4 - 3) \\ = 3 \times 2 \times 1 \\ = 6, \\ \\ the \: \: fifth \: \: term, \\ t _{5} = (5 - 1)(5 - 2)(5 - 3) \\ = 4 \times 3 \times 2 \\ = 24 \\ \\ In \: \: this \: \: way,\:\:we \: \: can \: \: find \: \: more \\ terms \: \: for \: \: the \: \: given \\ sequence \: \: and \: \: the \: \: next \: \: terms \\ be \: \: positive. \\ \\ Hence, \: \: PROVED. \\ \\ Thank \: \: you \: \: for \: \: the \: \: question.$