In the end, remainder matters

The remainder that a number leaves when divided by 9 is congruent to its digit sum modulo 9. (Refer to proof if necessary) Hence, the answer is

$\sum_{n=1}^{48}(2n)^2\equiv4\sum_{n=1}^{48}n^2\equiv4\times48\times49\times97\div6\equiv5\mod{9}$

Proof that the remainder that a number leaves when divided by 9 is congruent to its digit sum modulo 9:

Suppose we have a number $p=\overline{x_nx_{n-1}\cdots x_1}$ . Then,

$p\equiv x_n\cdot10^{n-1}+x_{n-1}\cdot10^{n-2}...+x_1\equiv x_n+...+x_1(\text{since any power of ten leaves a remainder of 1 when divided by 9})$ Thus, the proposition is proven.