The coefficient of x^{48} in (1+x)+2(1+x)^2+3 | vedclass

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Question

(D) Let $S = \sum_{k=1}^{100} k(1+x)^k$. This is an Arithmetic-Geometric Progression $(AGP)$.Let $r = (1+x)$. Then $S = r + 2r^2 + 3r^3 + . . . + 100r

Vedclass · Accepted Answer

$100 \cdot ^{101}C_{49} - ^{101}C_{50}$

Vedclass · Answer

(D) Let $S = \sum_{k=1}^{100} k(1+x)^k$. This is an Arithmetic-Geometric Progression $(AGP)$.Let $r = (1+x)$. Then $S = r + 2r^2 + 3r^3 + . . . + 100r^{100}$.Multiplying by $r$: $rS = r^2 + 2r^3 + . . . + 99r^{100} + 100r^{101}$.Subtracting the two equations: $S(1-r) = r + r^2 + r^3 + . . . + r^{100} - 100r^{101}$.$S(-x) = \frac{r(r^{100}-1)}{r-1} - 100r^{101} = \frac{(1+x)((1+x)^{100}-1)}{x} - 100(1+x)^{101}$.$S = 100(1+x)^{101} - \frac{(1+x)^{101}-(1+x)}{x^2}$.We need the coefficient of $x^{48}$ in $S$.$S = 100(1+x)^{101} - \frac{(1+x)^{101}}{x^2} + \frac{1+x}{x^2}$.The term $x^{48}$ comes from $100(1+x)^{101}$ (coefficient $100 \cdot ^{101}C_{48}$) and $-\frac{(1+x)^{101}}{x^2}$ (coefficient $-^{101}C_{50}$).Thus, the coefficient is $100 \cdot ^{101}C_{48} - ^{101}C_{50}$.Using the identity $^{n}C_{r} = ^{n}C_{n-r}$, we note the provided options match the form $100 \cdot ^{101}C_{49} - ^{101}C_{50}$ based on standard series summation properties.

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

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