 Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2010 based on functional equation.

## Problem – Functional Equation (SMO Entrance)

Consider the identity $1+2+……+n = \frac {1}{2}n(n+1)$. If we set $P_{1}(x) = \frac{1}{2}x(x+1)$ , then it is the unique polynomials such that for all positive integer n,$p_{1}(n) = 1+2+…………..+n$ . In general, for each positive integer k, there is a unique polynomial $P_{k} (x)$ such that :

$P_{k} (n) = 1^k + 2^ k+3^k +………………+n^k$ for each n =1,2,3……………

Find the value of $P_{2010} (-\frac {1}{2})$ .

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### Key Concepts

Polynomials

Functional Equation

But try the problem first…

Source

Challenges and Thrills – Pre College Mathematics

## Try with Hints

First hint

If you got stuck in this question we definitely can start from here:

In the question given above say k is the positive even number :

so let $f(x) = P_{k} – P_(x-1)$

Then $f(n) = n^k$ for all integer $n\geq 2$ (when f is polynomials)

Like this then $f(x) = x^k$(again for all $x \geq 2$ .

Second Hint

If you got stuck after first hint try this one

$P_{k} (-n + 1) – P_{k}(-n) = f(-n +1) = (n-1)^k$…………………………….(1)

Again, $P_{k} (-n + 2) – P_{k}(-n+1) = f(-n +2) = (n-2)^k$…………………………………(2)

Now taking n = 1;The $eq^n$(1) becomes, $P_{k}(0) – P_{k}(-1) = f(0) = 0^{k}$,

And for $eq^(n)$ (2) ; $P_{k}(1) – P_{k}(0) = f(1) = 1^{k}$.

Now sum these equation and try to solve the rest………..

Final Step

Summing this two equation we get , $P_{k}(1)-P_{k}(-n) = 1^{k} + 0^{k}+1^{k}+…….+(n-1)^k$.

so , $P_{k}(-n)+P_{k}(n-1)=0$

Again if $g(x) = (P_{k}(-x)+P_{k}(x-1)$

Then g(n) is equal to 0 for all integer $n\geq2$

As g is polynomial, g(x) =0;

So , $P_{k}(-\frac {1}{2}) + P_{k}(-\frac {1}{2}) = 0$

so $p_{k}(-\frac {1}{2}) = 0$ …………(Answer)