Space and time complexity of \$L = \{a^nb^{n^2} \mid n≥1\}\$

Consider the following language: $$L = \{a^nb^{n^2} \mid n≥1\}\,$$

When it comes to determining time and space complexity of a multi-tape TM, we can use two memory tapes, the first one to count $$n$$, and the second one to repeat $$n$$ times the count of $$n$$. Thus, because of the way we’re using the second tape, it should have a $$\Theta(n^2)$$ space complexity, and I would say the same concerning the time one. I thought it was correct, but the solution is $$TM(x)=|x|+n+2$$, where, $$x$$ is, supposedly, the length of the string, hence $$\Theta(|x|)$$. It seems correct to me, so is my reasoning completely wrong, or just a different way to express it?

Could we have reasoned about it differently, and say, for example, for every $$a$$ we write down a symbol on the first tape, and then count the $$b$$‘s, by scanning the symbols back and forth $$n$$ times? This time, the space complexity should just be $$\Theta(n)$$, while the time complexity should remain unchanged. What would change if we had a single-tape TM?