$\sum_{i=0}^{50}sgn((2i-1)(2i-3)\ldots(2i-99))$
There are 50(even number) terms present in the product.
Now we need to open summation for $i = 0$ to $50$
$i=0 , sgn((-1)(-3)\ldots(-99)) = +1$
Because we know that there are $50$ terms present in product which is an even number so it will be $+1.$
$i=1, sgn((1)(-1)\ldots (-97)) = -1$
Because in product there are $49$ negative terms and $1$ term is positive, answer will be negative.
From $i=0$ and $i=1,$ we understood that summation is giving $+1$ for even values of $i$ and $-1$ for odd values of $i.$
From $0$ to $50$ there are $26$ even and $25$ odd numbers are present.
So, answer will be $+1.$