Mildly Insane

$\begin{aligned} f(x) & = \sum_{i=1}^{50} {(x+2i-1)(x+2i)} \\ & = \sum_{i=1}^{50} {\left(x^2 +(4i-1)x+(4i^2-2i) \right)} \\ & = \left( \sum_{i=1}^{50} {1}\right) x^2 + \left(4 \sum_{i=1}^{50} {i} - \sum_{i=1}^{50} {1} \right)x + 4 \sum_{i=1}^{50} {i^2} - 2\sum_{i=1}^{50} {i} \\ & = 50x^2 + \left(4 \left[ \frac{(50)(51)}{2}\right]- 50 \right)x + 4 \left(\frac{(50)(51)(101)}{6}\right) - 2\left(\frac{(50)(51)}{2} \right) \\ & = 50x^2 + 5050x + 169150 \end{aligned}$

Using Vieta's formulas, we note that:

$S= - \dfrac{5050}{50} = - 101 \space$ and $\space P = \dfrac{169150}{50} = 3383$

$\Rightarrow \lfloor S+P \rfloor = \lfloor -101+3383 \rfloor = \boxed{3282}$

If we expand the summation , we get, $f (x) = (x+1)(x+2) + (x+3)(x+4) + (x+5)(x+6) + \cdots \cdots + (x+99)(x+100)$ Little bit of simplifying and we get, $f (x) = 50x^2 + x ( 1+2+3+ \cdots \cdots + 99+100 ) + ( 1\times 2 + 3\times 4 + 5\times 6 + \cdots \cdots + 99\times 100 )$ So, sum of the roots is negative of coefficient of $x$ divided by coefficient of $x^2$ , $S = - \dfrac{1+ 2+3+4+ \cdots \cdots +99+100 }{50}$ $= - \dfrac{\displaystyle\sum_{n=1}^{100} n }{50}$ $= -101$ And product is equal to coefficient of term independent of $x$ divided by coefficient of $x^2$ , $P = \dfrac{1\times 2 + 3\times 4 + \cdots \cdots + 99\times 100 }{50}$ $= \dfrac{ \displaystyle\sum_{n=1}^{50} (2n-1)(2n)}{50}$ $= \dfrac{ \displaystyle\sum_{n=1}^{50} 4n^2 - 2n}{50}$ $= \dfrac{ 4 \displaystyle\sum_{n=1}^{50} n^2 - 2\displaystyle\sum_{n=1}^{50} n }{50}$ $= \dfrac{ \dfrac{ 4 (50)(51)(101)}{6} - \dfrac{ 2(50)(51)}{2} }{50}$ $= 3383$ Now , $S = -101$ and $P= 3383$ . Addind them and taking its floor , $\lfloor S+P \rfloor =\lfloor (-101) + 3383 \rfloor = \boxed{3282}$