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1975 USAMO Problems/Problem 3

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Problem

If P(x) denotes a polynomial of degree n such that P(k)=k/(k+1) for k=0,1,2,\ldots,n, determine P(n+1).

Solution

Let Q(x) = (x+1)P(x) - x. Clearly, Q(x) has a degree of n+1.

Then, for k=0,1,2,\ldots,n, Q(k) = (k+1)P(k) - k = (k+1)\dfrac{k}{k+1} - k = 0.

Thus, k=0,1,2,\ldots,n are the roots of Q(x).

Since these are all n+1 of the roots, we can write Q(x) as: Q(x) = K(x)(x-1)(x-2) \cdots (x-n) where K is a constant.

Thus, (x+1)P(x) - x = K(x)(x-1)(x-2) \cdots (x-n)

Plugging in x = -1 gives:

(-1+1)P(-1) - (-1) = K(-1)(-1-1)(-1-2) \cdots (-1-n)

1 = K(-1)^{n+1}(1)(2) \cdots (n+1)

K = (-1)^{n+1}\dfrac{1}{(n+1)!}

Finally, plugging in x = n+1 gives:

(n+1+1)P(n+1) - (n+1) = (-1)^{n+1}\dfrac{1}{(n+1)!}(n+1)(n+1-1)(n+1-2) \cdots (n+1-n)

(n+2)P(n+1) = (-1)^{n+1}\dfrac{1}{(n+1)!}\cdot(n+1)! + (n+1)

(n+2)P(n+1) = (-1)^{n+1} + (n+1)

P(n+1) = \dfrac{(-1)^{n+1} + (n+1)}{n+2}

If n is even, this simplifies to P(n+1) = \dfrac{n}{n+2}. If n is odd, this simplifies to P(n+1) = 1.

See also

1975 USAMO (Problems)
Preceded by
Problem 2
1 2 3 4 5 Followed by
Problem 4
All USAMO Problems and Solutions
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