Let $ Q(x)=(x+1)P(x)-x$ (eq 1) Then $ Q(x)$ is a polynomial of degree $ n+1$ Now for all integers $ i$ in the range $ 0\le i\le n$ $ Q(i)=0$ So by the Factor Theorem: $ Q(x)=(x-0)(x-1)...(x-n).R(x)$ where $ R(x)$ is a polynomial $ deg(R)=deg(Q)-deg((x-0)(x-1)...(x-n))=(n+1)-(n+1)=0$ So $ R$ is a constant polynomial. Let $ R(x)=k$ Then $ Q(x)=kx(x-1)...(x-n)$ (eq 2) x=-1 in (eq 1) gives $ Q(-1)=1$ x=-1 in (eq 2) gives $ Q(-1)=k(-1)(-2)...(-n-1)=k (-1)^{n+1} [(n+1)!]$ So $ k (-1)^{n+1} [(n+1)!]=1$ So $ k [(n+1)]!=(-1)^{n+1}$ x=n+1 in (eq 2) gives $ Q(n+1)=k [(n+1)!]$ But $ k [(n+1)]!=(-1)^{n+1}$ So $ Q(n+1)=(-1)^{n+1}$ Substitute this in (eq 1) gives $ (n+2)P(n+1)=(n+1)+(-1)^{n+1}$ So when $ n$ is even $ P(n+1)=\frac{n}{n+2}$ And when $ n$ is odd $ P(n+1)=1$

Dang, USAMO was easy way back then.

I may be high right now but I just cant seem to follow this step, tried for about 30min $=(-1)^{n+1}C(n+1)!=1 $ then $ C=\frac{(-1)^{n+1}}{(n+1)!} $ why is this true o.o so messed

oh woww, didnt even realize that :O ty

Let \[Q(x)=P(x)(x+1)-x \qquad (1).\]Since $ P(k)=\frac{k}{k+1}$ for $ k=0,1,2,\ldots,n$, it is easy to see that the roots of $Q(x)$ are $0,1,\ldots n$, so we can write \[Q(x)=Mx(x-1)(x-2) \ldots (x-n) \qquad (2)\]for some number $M$. Taking $x=-1$ in both equation (1) and (2) gives $Q(-1)=1$ and $Q(-1)=M(-1)^{n+1}(n+1)!$, respectively. Equating them, we get that \[1=M(-1)^{n+1}(n+1)! \]\[M(n+1)!=(-1)^{n+1} \qquad (3).\]Further, taking $x=n+1$ in (2) gives \[Q(n+1)=M(n+1)! \qquad (4).\]Equating (3) and (4), $Q(n+1)=(-1)^{n+1}.$ Recall that $Q(x)=P(x)(x+1)-x$, so we have that $P(n+1)(n+2)-n-1=(-1)^{n+1}$. Hence, \[P(n+1)=\frac{(-1)^{n+1}+n+1}{n+2}.\]If $n$ is even the RHS is $\frac{n}{n+2}$ and if $n$ is odd the RHS is $1$.

Consider the polynomial \[Q(x) = (x+1)\cdot P(k) -k\]Note that for $x=0,1,2,\ldots, n$ we have $Q(x)=0$. Thus, we may write \[Q(x) = c\cdot \prod_{i=0}^n (x-i)\]Then, answer extract \[P(n+1) = \frac{Q(n+1)+n+1}{n+2} = \frac{(-1)^{n+1}\cdot Q(-1)+n+1}{n+2} = \frac{(-1)^{n+1}+n+1}{n+2}\]

Brut3Forc3 wrote: If $ P(x)$ denotes a polynomial of degree $ n$ such that $ P(k)=\frac{k}{k+1}$ for $ k=0,1,2,\ldots,n$, determine $ P(n+1)$. Define $$Q(x):=(x+1)\cdot P(x) -x$$Since for $x \in \{1,2,......,k\}$ we must have $Q(x)=0$ it follows that $$Q(x) = c\cdot \prod_{i=0}^n (x-i)$$hence $$P(n+1) = \frac{Q(n+1)+n+1}{n+2} = \frac{(-1)^{n+1}\cdot Q(-1)+n+1}{n+2} = \frac{(-1)^{n+1}+n+1}{n+2}\;\;\;.\blacksquare$$

Define the polynomial $Q(x)=x+1\cdot P(x)-x.$ Notice $Q$ has roots at $x=0,1,2,\ldots, n.$ Thus we can write $Q(x)=cx(x-1)(x-2)\cdots(x-n)$ for some constant $c.$ Considering $Q(-1),$ we notice $Q(-1)=1,$ However, we can alternatively express \[Q(-1)=c(-1)(-2)(-3)\cdots(-1-n) = c(-1)^{n+1}(n+1)! \implies c = \frac{(-1)^{n+1}}{(n+1)!}.\]It follows $Q(n+1)=\tfrac{(-1)^{n+1}}{(n+1)!}n+1(n)(n-1)\cdots1= \tfrac{(-1)^{n+1}}{(n+1)!} (n+1)!=(-1)^{n+1},$ so \[P(n+1) = \frac{Q(n+1)+n+1}{n+2} = \frac{(-1)^{n+1}+n+1}{n+2},\]as required. Remark. I wonder if there is a possible solution directly interpolating $P(n+1) = \textstyle\sum_{i=0}^n \tfrac{i}{i+1} (-1)^i \tbinom{n+1}{i}.$

Given $P(x)$ is a polynomial of degree $n$ and also given $P(i)=\frac{i}{i+1}$ for $i=0,1,2,3,4,5......,n$ we have to find the value of $P(n+1)$ $$P(i)=\frac{i}{i+1}\implies(i+1)P(i)-i=k\prod_{i=0}^{n}(x-i)$$Putting $-1$ we get, $1=k(-1)(-2)....(-1-n)\implies \boxed{k=\frac{-1}{(n+1)!}}$ or $\boxed{k=\frac{1}{(n+1)!}}$ When $n$ is even then we get, $$(x+1)P(x)-x=\frac{(x)(x-1)(x-2)(x-3)(x-4)...(x-n)}{(n+1)!}$$Putting $n+1$ get, $$(n+2)P(n+1)-(n+1)=\frac{(n+1)(n)....(1)}{(n+1)!}\implies \boxed{P(n+1)=1}$$When $n$ is odd get, $$(x+1)P(x)-x=-\left[\frac{(x)(x-1)....(x-n)}{(n+1)!}\right]$$Putting $n+1$ get, $$(n+2)P(n+1)-(n+1)=-\left[\frac{(n+1)!}{(n+1)!}\right]\implies\boxed{P(n+1)=\frac{n}{n+2}}$$

$P(k)=\frac{\pm\frac{k(k-1)(k-2)...(k-n)}{(n+1)!}+k}{k+1}$, so the answer is $1$ for odd $n$ and $\frac{n}{n+2}$ for even $n$

Given:- p(x) is a polynomial of degree n such that p(k) = k/k + 1 for k = 0,1,2,.......,n. (k + 1)p(k) - k = 0 for k = 0,1,2,.......,n (k + 1)p(k) - k = a(k)(k - 1)(k - 2)........(k - n) where a is the leading coefficient. On plugging k = -1 in the above equation to eliminate the (k + 1).p(k) term and get the value of the leading coefficient a, we get 1 = a(-1)(-2)(-3)........(-1 - n) a[(-1)^n + 1 (n + 1)!] = 1 a = (-1)^n + 1/(n + 1)! Now, On plugging k = n + 1 in the same equation, we get (n + 2).p(n + 1) -(n + 1) = a[(n + 1).n.(n - 1).........1] (n + 2).p(n + 1) -(n + 1) = a(n + 1)! (n + 2).p(n + 1) -(n + 1) = (-1)^n + 1 (n + 2).p(n + 1) = (-1)^n + 1 + (n + 1) Therefore, p(n + 1) = [(-1)^n + 1 + (n + 1)]/(n + 2) If n is odd then p(n + 1) = 1 and if n is even then p(n + 1) = n/n + 2

here a solution using interpolation (it is of course not needed, I solved it in the same way as everyone else at first, but at least it is a nice problem to practice some binomial coefficient manipulation)