Find $p(n+2)$ where $p(x)=\sum_{i=1}^{n+1}2^{i}\cdot\frac{\prod_{j=1,j\neq i}^{n+1}(x-j)}{\prod_{j=1,j\neq i}^{n+1}(i-j)}$.

Question

Define $$p(x)= \sum_{i=1}^{n+1}2^{i}\cdot\frac{\prod_{j=1,j\neq i}^{n+1}(x-j)}{\prod_{j=1,j\neq i}^{n+1}(i-j)}.$$

Find $p(n+2)$.

The polynomial is rather complicated. I tried but can not succeed.

According to my teacher, the result is $2^{n+2}-1$.

But how we can prove it.

Answer 1

Show that $p$ is of degree at most $n$ and that $p(i)=2^i$ for $i=1,2,\ldots,n+1$. Let $q(x):=2 \,\sum_{r=0}^n\,\binom{x-1}{r}$. Prove that $p=q$ and $q(n+2)=2^{n+2}-2$. Either you copied the problem incorrectly or your teacher is wrong.

Answer 2

By Lagrange interpolation, $p(x)$ is a polynomial with degree $n$ that equals $2^i$ at $x=i$, for any $i\in [1,n+1]$. Let $\delta$ be the difference operator that maps a polynomial $q(x)$ into $q(x+1)-q(x)$. $\delta p$ is a polynomial with degree $n-1$ that equals $2^i$ at $x=i$, for any $i\in[1,n]$. By iterating the argument, $\delta^n p$ is a polynomial with degree zero, i.e. a constant polynomial, that equals $2$ at $x=1$. It follows that $\delta^n p$ equals $2$ at any point: in particular, $(\delta^n p)(2)=2$, hence

$$(\delta^{n-1} p)(3) = (\delta^{n-1}p)(2)+(\delta^n p)(2) = 2^2+2 = 6,$$ $$(\delta^{n-2} p)(4) = (\delta^{n-2}p)(3)+(\delta^{n-1} p)(3) = 2^3 + 6 = 14,$$ $$ \ldots $$ $$(\delta^{n-k}p)(k+2) = 2^{k+1}+2^{k}+\ldots+2^{1} = 2^{k+2}-2 $$ and by taking $k=n$ in the last identity, the correct claim $\color{red}{p(n+2)=2^{n+2}-2}$ follows.