Let $r=1+x$ and $S={\sum }_{k=1}^{100}k{r}^k$. This is an arithmetico-geometric progression; multiply by $r$ and subtract:

$S-rS=(r+r^2+\dots +r^{100})-100r^{101}$.

Solving with $r-1=x$ gives $S=-\frac{(1+x{)}^{101}}{x^2}+\frac{1}{x^2}+\frac{100(1+x{)}^{101}}{x}$.

Extract coefficient of $x^{48}$:

From $\frac{100(1+x{)}^{101}}{x}$: equals 100 $\times$ (coefficient of $x^{49}$ in $(1+x{)}^{101}$) $=100\cdot {}^{101}{C}_{49}$.

From $-\frac{(1+x{)}^{101}}{x^2}$: equals $-1\times$ (coefficient of $x^{50}$ in $(1+x{)}^{101}$) $=-{}^{101}{C}_{50}$.

The $\frac{1}{x^2}$ term has only the $x^{-2}$ contribution, nothing at $x^{48}$.

Total: $100\cdot {}^{101}{C}_{49}-{}^{101}{C}_{50}$.