Christmas Streak 32/88: Polynomial But Strange

Employ the fact that $P(x+1)-P(x)$ decreases the degree of a polynomial.

Let's calculate with $\begin{array}{ccc} a&&b \\ &b-a&\end{array}$

Then

$\begin{array}{ccccccccc} P(1)=\dfrac{1}{3}&&P(2)=\dfrac{1}{7}&&P(3)=\dfrac{1}{13}&&P(4)=\dfrac{1}{21}&&P(5)=x \\ &-\dfrac{4}{21}&&-\dfrac{6}{91}&&\dfrac{-8}{273}&&x-\dfrac{1}{21}& \\ &&\dfrac{34}{273}&&\dfrac{10}{273}&&x-\dfrac{5}{273}&& \\ &&&-\dfrac{24}{273}&&x-\dfrac{15}{273}&&& \\ \end{array}$

Therefore $P(5)=x=-\dfrac{9}{273}=-\dfrac{3}{91}.$

$\therefore~a+b=\boxed{94}.$

Here is a different solution that I expect most people would have used;

Let the unknown cubic polynomial be of the form

$C_1 x^3 + C_2 x^2+C_3 x+C_4$

Using the next piece of information given, that when $x=1,2,3,4, P(x)=\large \frac{1}{1+x+x^2}$ , we can write the following equations:

For $x=1$ ;

$C_1+C_2+C_3+C_4=\large \frac{1}{1+1+1^2}$

For $x=2$ ;

$8C_1+4C_2+2C_3+C_4=\large \frac{1}{1+2+2^2}$

For $x=3$ ;

$27C_1+9C_2+3C_3+C_4=\large \frac{1}{1+3+3^2}$

And for $x=4$ ;

$64C_1+16C_2+4C_3+C_4=\large \frac{1}{1+4+4^2}$ .

There we have it. 4 unique equations for 4 variables. We can now find the value of the coefficients of the cubic function by solving the system of linear equations above.

The solutions are;

$C_1=\large \frac{-4}{273} , C_2 = \large \frac{41}{273} , C_3 = \large \frac{-7}{13} , C_4 = \large \frac{67}{91}$

If we find the value of $P(5)$ for $P(x) = \large \frac{-4}{273} x^3 + \large \frac{41}{273} x^2 + \large \frac{-7}{13} x + \large \frac{67}{91}$ , we get:

$\large \frac{-3}{91}$

Therefore $a+b=3+91=\large \boxed{94}$