If $p(x)$ is a polynomial of degree n such that $p(x)=2^x$ for $x=1,2,3,4,5,...n+1$ . Find $p(x+2)$ and $p(x+7)$

Question

If $p(x)$ is a polynomial of degree n such that $p(x)=2^x$ for $x=1,2,3,4,5,...n+1$ . Find $p(x+2)$ and $p(x+7)$. x I solve this problem by replacing $x$ by $x+2$ in $p(x)$,

I got $p(x+2)=2^{x+2}$ and similarly I got $p(x+7)=2^{x+7}$

Is my method to find answer is correct? If not, please add solution to this question.

Answer 1

Hint. Note that for $j=1,\dots, n+1$, $$P_n(j)=2^j=2\cdot(1+1)^{j-1}=2\sum_{k=0}^{j-1}\binom{j-1}{k}=2\sum_{k=0}^n\binom{j-1}{k}$$ where $$\binom{x}{k}:=\frac{x(x-1)\cdots (x-k+1)}{k!}.$$ I guess that you are looking for the values of $P_n(n+2)$ and $P_n(n+7)$. Can you take it from here?