<s>[INST] Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\frac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$, find the value of $P(n+1)$ Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\dfrac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$
I tried by making a $f(x) = (x^3+1) P(x) - x^2$. But this equation will have $n+3$ roots, out of which $n+1$ will be $0,1,2,...,n$. What about the other two roots? Can we simplify $x^2 \over x^3+1$ more to help?
 [/INST] Using (Lagrange) polynomial interpolation:
\begin{align*}
P(x) &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \prod_{j \neq k} \frac{x - j}{k-j} \right) \\
&= \sum_{k=0}^n \frac{k^2}{k^3+1} \left(  \frac{x (x-1) \cdots (x - (k-1)) (x-(k+1)) \cdots (x-n) }{ k (k-1) \cdots (k-(k-1)) (k-(k+1)) \cdots (k-n)}\right) \\
&= \sum_{k=0}^n \frac{k^2}{k^3+1} \left(  \frac{x (x-1) \cdots (x - (k-1)) (x-(k+1)) \cdots (x-n) }{ k! (n-k)!(-1)^{n-k} } \right) \\
\end{align*}
Substituting,
\begin{align*}
P(n+1) &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left(  \frac{(n+1) (n) \cdots (n+2-k) (n-k) \cdots (1) }{ k! (n-k)!(-1)^{n-k} } \right) \\ 
&=  \sum_{k=0}^n \frac{k^2}{k^3+1} \left(  \frac{(n+1)!}{ k! (n-k)!(-1)^{n-k} (n+1-k)} \right) \\ 
&=  \sum_{k=0}^n (-1)^{k-n} \frac{k^2}{k^3+1} {n+1 \choose k} \\
\end{align*}
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