Finding a value from 5 systems of equations of 5 variables(CHMMC 2014) $$\text{For } a_1\cdots a_5\in \mathbb{R},$$
$$\frac{a_1}{k^2+1}+\cdots+\frac{a_5}{k^2+5}=\frac{1}{k^2}$$
$$\forall k=\{2,3,4,5,6\}$$
$$\text{Find }\frac{a_1}2+\cdots+\frac{a_5}6$$
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 $a_1\cdots a_5$.

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.
