Question about $T_n = n + (n-1)(n-2)(n-3)(n-4)$

Question

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?

Answer 1

Your reasoning is correct. You can generalize it to say that

$$T_n = (an + b) + (n-1)(n-2)\cdots(n-k)$$

is an arithmetic progression for $1 \le n \le k$, for some arbitrary integer $k \ge 1$. This is because similarly, the product evaluates to $0$ for these values of $n$, to leave $$T_n = an + b$$

This remainder essentially defines an arithmetic progression with common difference $a$.

Answer 2

$d=T_{n}-T_{n-1}$ $= n + (n-1)(n-2)(n-3)(n-4)- (n-1 )- (n-2)(n-3)(n-4)(n-5)$

$T_{n}-T_{n-1}=1+(n-2)(n-3)(n-4)(n-1-(n-5))=1+4(n-2)(n-3)(n-4)$

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

Answer 3

By distributing only the first term of each parentheses in the product, we get the term $n^4$. There is no other way to obtain such a term. Hence $T(n)$ is a 4th order polynomial and 4th order polynomials never give arithmetic sequences.