I don’t understand how to reduce this fraction to the stated solution: The fraction is as follows:
$$
\frac{9 \cdot 11 + 18 \cdot 22 + 27 \cdot 33 + 36 \cdot 44 }{
22 \cdot 27 + 44 \cdot 54 + 66 \cdot 81 + 88 \cdot 108}
$$
That’s all fine. What I don’t get is that my textbook says this reduces to the following:
$$\frac{9\times 11 + (1^2 + 2^2 + 3^2 + 4^2)}{22 \times 27 \times (1^2 + 2^2 + 3^2 +4^2)}$$
I don’t understand how the sum of consecutive squares can be deduced from that fraction, or why the denominator contains $22\times 27 \times\dots $ as opposed to the numerator which is $9 \times 11 + \dots$”
Any insight would be really appreciated!
 A: First correct the error in your expression. The first addition sign on top should be a multiplication sign.
Observe for instance that on the numerator, you have the first term $9\cdot11= (9\cdot 1)(11\cdot 1) = 9\cdot11(1^2) $  and the second term is $18\cdot 22 = (9\cdot 2)(11\cdot 2) = 9\cdot11 (2^2)$.
Using exactly the same reasoning, you can convert the remaining terms on top to very similar forms, allowing you to express the numerator as $9\cdot11(1^2 + 2^2 + 3^2 + 4^2)$.
Apply analogous reasoning to the denominator (here you're dealing with terms like $22\cdot27(1^2),22\cdot27(2^2)\dots$ etc.
A: There seems to be a mistake: 
\begin{align*}
\frac{9\cdot 11 + 18 \cdot 22 + 27 \cdot 33 + 36 \cdot 44}{22\cdot 27 + 44\cdot 54 + 66 \cdot 81 + 88 \cdot 108} & = \frac{9\cdot 11(1\cdot 1+2\cdot 2+3\cdot 3 + 4\cdot 4)}{22\cdot 27(1\cdot 1+2\cdot 2+3\cdot 3+4\cdot 4)} \\
& = \frac{9\cdot 11(1^2+2^2+3^2 + 4^2)}{22\cdot 27(1^2+2^2+3^2+4^2)} = \frac{99}{594} < 1,
\end{align*}
and this is different from $99+\frac{1}{594} > 99$, which is the result you have.
