Show that $-1^2+3^2-5^2\mp ...+(2^n-1)^2=2^{2n-1}$ Playing with numbers, I construct following expression.

Can it be shown that
$$\sum_{i=1}^{2^{n-1}}(-1)^i(2i-1)^2=2^{2n-1}$$

attempt
We can construct following, using finite calculus as
$(-1)^2+3^2+7^2+...+(4k-5)^2 = \binom{k}1+8\binom{k}2+32\binom{k}3\quad\quad eq(1)$
$1^2+5^2+9^2+...+(4k-3)^2 = \binom{k}1+24\binom{k}2+32\binom{k}3\quad\quad eq(2)$
Let $4k-1=2^n-1$ so we can write above claim as
$eq(1)-eq(2)-1+(4k-1)^2=2^{2n-1}$
Which is equivalent to show
$(2^n-1)^2-16\binom{2^{n-2}}2-1= 2^{2n-1}$
Here I'm stuck. Thanks

Using the Hockey-stick identity:
$$S(n)=\sum_{k=0}^{n-1} (2k+1)^2=\sum_{k=0}^{n-1} \left(8{k+1\choose 2}+1\right)=8{n+1\choose 3}+n=\frac{n(2n+1)(2n-1)}{3}$$
$$T(n)=\sum_{k=0}^{n-1} (4k+1)^2=\sum_{k=0}^{n-1} \left( 16k^2+8k+1\right)=\sum_{k=0}^{n-1} \left(32{k+1\choose2}-8{k+1\choose1}+9\right)\\=32{n+1\choose3}-8{n+1\choose2}+9n=\frac n3(16n^2-12n-1)$$
Let $N=2^{n-2}$, then
$$-1^2+3^2-5^2+\dots+(2^n-1)^2\\=-1^2+3^2-5^2+\dots+(4N-1)^2\\=S(2N)-2T(N)\\=8N^2\\=2^{2n-1}$$

To answer the comment:
$$1^2+3^2+5^2+\dots+(4N-1)^2=\sum_{k=0}^{2N-1} (2k+1)^2=S(2N)$$
$$1^2+5^2+\dots+(4N-3)^3=1^2+5^2+\dots+(4(N-1)+1)^3=\sum_{k=0}^{N-1}(4k+1)^2=T(N)$$
Now, the first row minus twice the second row yields:
$$-1^2+3^2-5^2+\dots+(4N-1)^2=S(2N)-2T(N)$$
