Finding a value from 5 systems of equations of 5 variables(CHMMC 2014)

Question

$$\text{For } a_1\cdots a_5\in \mathbb{R},$$ $$\frac{a_1}{k^2+1}+\cdots+\frac{a_5}{k^2+5}=\frac{1}{k^2}$$ $$\forall k=\{2,3,4,5,6\}$$ $$\text{Find }\frac{a_1}2+\cdots+\frac{a_5}6$$ The Provided explanation/official solution was $$\text{Solution }1:\large{\frac{65}{72}}$$ Please don't mark this as off-topic, I have no clue where to start. Clearly it would not be feasible to compute $a_1\cdots a_5$.

Answer 1

Let $$\sum_{i=1}^{5} \frac{a_i}{x+i} = \frac{P(x)}{Q(x)}$$ where $P(x)$ is a polynomial of degree $4$, and $Q$ of degree $5$. The coefficent of $x^4$ in $P(x)$ is $\sum a_i$ and $Q(x) = \prod_{k=1}^{5} (x+k)$

We get that

$$x P(x) - Q(x) = 0$$

as $0 = \dfrac{P(x)}{Q(x)} - \dfrac{1}{x} = \dfrac{x P(x) - Q(x)}{xQ(x)}$ for $x = 2^2, 3^2, \dots, 6^2$.

for $x = 2^2, 3^2, \dots, 6^2$ Thus $$xP(x) - Q(x) = K \prod_{k=2}^6 (x - k^2)$$

(as $xP(x) - Q(x)$ is a polynomial of degree $5$, and leading coefficient $\sum a_i -1$)

We are looking at $$\frac{P(1)}{Q(1)} = \frac{Q(1) + K\prod_{k=2}^6 (1 - k^2)}{Q(1)}$$

Now $Q(1) = 6! = 720$ And $\prod_{k=2}^{6} (1-k^2) = -302400$

Now we need to figure out that value $K$. This we can, by setting $x=0$ in $$xP(x) - Q(x) = K \prod_{k=2}^6 (x - k^2)$$

Answer 2

Starting from $\displaystyle \frac{a_1}{k^2+1}+\cdots+\frac{a_5}{k^2+5}=\frac{1}{k^2}$ we would like to plug in $k^2=1$ and get

$$ \frac{a_1}{2}+\cdots+\frac{a_5}{6}=1$$

In fact, the correct answer is slightly less. We can add any function rational $f(k)$ which vanishes on all of $k^2 = 2,3,4,5,6$. So are are going to define $f(x) = K \cdot \prod_{k=2}^6 (x - k^2)$ where $g(x)$ can be any polynomial.

$$ \frac{a_1}{x^2+1}+\cdots+\frac{a_5}{x^2+5}=\frac{1+ K \cdot \prod_{k=2}^6 (x - k^2)}{x^2} \tag{$\ast$}$$

If we set $x= 0$ the right side is undefined, but we can cross multiply:

\begin{eqnarray} x^2 \left( \frac{a_1}{x^2+1}+\cdots+\frac{a_5}{x^2+5} \right)&=&1+ K \cdot \prod_{k=2}^6 (x - k^2) \\ 0&=&1 - K \cdot 6!^2 \end{eqnarray}

We have that $\boxed{K = \frac{1}{6!^2}}$ Then plug $x=1$ into $(\ast)$

$$ \frac{a_1}{2}+\cdots+\frac{a_5}{6}=\frac{1+ K \cdot \prod_{k=2}^6 (x - k^2)}{x^2} = 1 - \frac{(3\cdot 4 \cdot 5 \cdot 6 \cdot 7)(1\cdot 2 \cdot 3 \cdot 4 \cdot 5)}{6!^2 }$$

This argument is not quite seamless as the RHS is off by factor of 6.