Competency-Based
Assessment Made Simple.

Let $S = 1(1+x) + 2(1+x)^2 + 3(1+x)^3 + \dots + 100(1+x)^{100}$. This is an Arithmetico-Geometric Progression (AGP). Multiply by $(1+x)$: $(1+x)S = 1(1+x)^2 + 2(1+x)^3 + 3(1+x)^4 + \dots + 99(1+x)^{100} + 100(1+x)^{101}$. Subtracting the two equations: $S - (1+x)S = (1+x) + (1+x)^2 + (1+x)^3 + \dots + (1+x)^{100} - 100(1+x)^{101}$. $-xS = \frac{(1+x)( (1+x)^{100} - 1 )}{(1+x) - 1} - 100(1+x)^{101}$. $-xS = \frac{(1+x)( (1+x)^{100} - 1 )}{x} - 100(1+x)^{101}$. $S = -\frac{(1+x)( (1+x)^{100} - 1 )}{x^2} + \frac{100(1+x)^{101}}{x}$. $S = \frac{100x(1+x)^{101} - (1+x)^{101} + (1+x)}{x^2}$. $S = \frac{(100x-1)(1+x)^{101} + (1+x)}{x^2}$.

We need the coefficient of $x^{48}$ in $S$. This is equivalent to finding the coefficient of $x^{50}$ in $(100x-1)(1+x)^{101} + (1+x)$. Coefficient of $x^{50}$ in $(100x-1)(1+x)^{101}$ is $100 \times$ (coefficient of $x^{49}$ in $(1+x)^{101}$) $- 1 \times$ (coefficient of $x^{50}$ in $(1+x)^{101}$). This is $100 \times {101 \choose 49} - {101 \choose 50}$. Coefficient of $x^{50}$ in $(1+x)$ is 0. Therefore, the coefficient of $x^{50}$ in $(100x-1)(1+x)^{101} + (1+x)$ is $100 {101 \choose 49} - {101 \choose 50} + 0 = 100 {101 \choose 49} - {101 \choose 50}$. The coefficient of $x^{48}$ in $S$ is $100 {101 \choose 49} - {101 \choose 50}$.