Question about $T_n = n + (n-1)(n-2)(n-3)(n-4)$ 
The formula $T_n = n + (n-1)(n-2)(n-3)(n-4)$ will produce an arithmetic sequence for $n < 5$ but not for $n \ge 5$. Explain why.

I think it is because if n is less than five the term with multiplication will be equal to zero and the common difference will be one. If n  is greater than five, the term added will have multiplication and there will not be a common difference.
Is this correct?

$d=T_{n}-T_{n-1}$ $= n + (n-1)(n-2)(n-3)(n-4)- (n-1 )- (n-2)(n-3)(n-4)(n-5)$
$T_{n}-T_{n-1}=1+(n-2)(n-3)(n-4)(n-1-(n-5))=1+4(n-2)(n-3)(n-4)$
here its obvious that if $1<n<5$ we have a constant diffference(since all $n$ vanish)  ie $d=1$
but $n\ge5$ we will have a 
a  variable dependent difference,hence not linear
