You need to take the sum of the first $n$ odd squares, e.g.if $n=3$, then you need to add $1+9+25=35$. And, that does indeed equal $\frac{4}{3}n^3-\frac{1}{3}n$ for $n=3$: $\frac{4}{3}3^3-\frac{1}{3}3=\frac{4}{3}27-1=36-1=35$
Now, to prove that this is true in general using induction:
Base: $n=0$: $\frac{4}{3}0^3-\frac{1}{3}0=0-0=0$ which is indeed the sum of the first $0$ odd squares. Check!
Step: Assume that for some $k$ the sum of the first $k$ odd squares is $\frac{4}{3}k^3-\frac{1}{3}k$. Now let's consider the sum of the first $k+1$ odd squares, which is of course the sum of the first $k$ odd squares plus the $k+1$-th odd square, which is $(2k+1)^2$. So, by the inductive hypothesis the sum is $\frac{4}{3}k^3-\frac{1}{3}k+(2k+1)^2$, and now you need to verify that this does indeed equal $\frac{4}{3}(k+1)^3-\frac{1}{3}(k+1)$. Let's see:
$$\frac{4}{3}k^3-\frac{1}{3}k+(2k+1)^2=$$
$$\frac{4k^3-k+3(4k^2+4k+1)}{3}=$$
$$\frac{4k^3+12k^2+12k-k+3}{3}=$$
$$\frac{4k^3+12k^2+12k+4-k+3-4}{3}=$$
$$\frac{4(k^3+3k^2+3k+1)-(k+1)}{3}=$$
$$\frac{4(k+1)^3}{3}-\frac{1}{3}(k+1)$$
Best Answer
Hint: As you mentioned, you have $k = 2l+1$ for some $l$.
From here, note the difference of squares:
$$a^2-b^2 = (a+b)(a-b)$$
Let $a$ and $b$ be two consecutive integers and try to simplify.