$Q(n)-Q(n-1) = T(n)$ Prove that $Q(n)$ degree is $k+1$

Question

I was given this problem and I've been thinking a lot of time and still I have nothing.

$Q:ℕ↦ℕ$
$Q(n)-Q(n-1) = T(n)$

$T(n)$ degree is $k$
Prove that $Q(n)$ degree is $k+1$

any idea?
Thank you

Answer 1

Let $Q(n)$ be a $r^{th}$ degree polynomial.

Thus $Q(n)=a_0n^r+a_1n^{r-1}+a_2n^{r-2}+\ldots+a_{r-1}n+a_r$

which means that $Q(n-1)=a_0(n-1)^r+a_1(n-1)^{r-1}+a_2(n-1)^{r-2}+\ldots+a_{r-1}(n-1)+a_r$

Now in the expansion of $(n-1)^r$, the term with highest power, that is $n^r$ always have a coefficient of $1$. Hence $(n-1)^r$ can be written as $n^r-P_r(n)$ where $P_r(n)$ is some polynomial of degree $r-1$

Thus $a_0(n-1)^r=a_0n^r-a_0P_r(n)$, $\space a_1(n-1)^{r-1}=a_1n^{r-1}-a_1P_{r-1}(n)$ and so on

$\therefore Q(n-1)=Q(n)-T(n)$ where $T(n)=a_0P_r+a_1P_{r-1}+\ldots+a_{r-1}P_1$

Now it is evident that $T(n)$ would be of degree $(r-1)$ and also $Q(n)-Q(n-1)=T(n)$

We have shown here that if the degree of $Q(n)$ is $r$, then the degree of $T(n)=Q(n)-Q(n-1)$ is $(r-1)$.

The degree of $T(n)$ given here is $k$

$\therefore $ The degree of $Q(n)$ is $(k+1)$