From Back to Power 4

$100^4-99^4\ldots+2^4-1^4$ can be written as

$\displaystyle \sum_{k=1}^n ((2k)^4-(2k-1)^4)$

= $\displaystyle \sum_{k=1}^n (32k^3-24k^2+8k-1)$

= $32 \dfrac{n^2(n+1)^2}{4}-24\dfrac{(n(n+1)(2n+1)}{6}+8 \dfrac{n(n+1)}{2}-n$

= $8n^4+8n^3-n$

With $n=50$ , $8(50)^4+8(50)^3-50$

= $50999950$

$100^4-99^4\ldots+2^4-1^4$ $=-\sum_{n=1}^{100}n^4+2\sum_{n=1}^{50}2n^4$ $=32\sum_{n=1}^{50}n^4-\sum_{n=1}^{100}n^4$ $=32\left(\frac{50^5}{5}+\frac{50^4}{2}+\frac{50^3}{3}+\frac{50}{30}\right)-\left(\frac{100^5}{5}+\frac{100^4}{2}+\frac{100^3}{3}+\frac{100}{30}\right)$ $=\huge50999950$

$\begin{aligned} S & = {\color{#3D99F6}-1^4 + 2^4} {\color{#D61F06} - 3^4 + 4^4} - \cdots {\color{#3D99F6}-97^4 + 98^4} \color{#D61F06}-99^4 + 100^4 \\ & = \sum_{k=1}^n \left(-(2k-1)^4+(2k)^4\right) & \small \color{#3D99F6} \text{where }n = 50 \\ & = \sum_{k=1}^n \left(32k^3-24k^2+8k-1 \right) \\ & = 32\times \frac {n^2(n+1)^2}4 - 24 \times \frac {n(n+1)(2n+1)}6 + 8 \times \frac {n(n+1)}2 - n \\ & = 4n(n+1)(2n(n+1)-(2n+1) + 1) -n \\ & = 8n^4+8n^3 - n & \small \color{#3D99F6} \text{putting back }n = 50 \\ & = \boxed{50999950} \end{aligned}$