The coefficient of x^48 in (1+x)+2(1+x)^2+3(1+x)^3++100(1+x)^100 is equal to

Question

The coefficient of $x^{48}$ in $(1+x)+2(1+x)^2+3(1+x)^3+\ldots+100(1+x)^{100}$ is equal to

Accepted Answer

Let $S=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+100(1+x)^{100}$.

Multiplying by $(1+x)$: $(1+x)S = (1+x)^2+2(1+x)^3+\cdots+100(1+x)^{101}$.

Subtracting, $-xS = (1+x)+(1+x)^2+\cdots+(1+x)^{100} - 100(1+x)^{101} = (1+x)\dfrac{(1+x)^{100}-1}{x} - 100(1+x)^{101}$

So $$S=\frac{(1+x)-(1+x)^{101}}{x^2}+\frac{100(1+x)^{101}}{x}$$

The coefficient of $x^{48}$ picks up $-{}^{101}C_{50}$ from the first term (coefficient of $x^{50}$ in $(1+x)^{101}$) and $100\cdot{}^{101}C_{49}$ from the second term (coefficient of $x^{49}$ in $(1+x)^{101}$). Hence

$$[\,x^{48}\,]S = 100\cdot{}^{101}C_{49} - {}^{101}C_{50}$$

Solutions · 1