Progression (Sequence and Series)Medium
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 :
Options
A.$100.\ ^{100}C_{49} - \ ^{100}C_{50}$
B.$\ ^{100}C_{50} + \ ^{101}C_{49}$
C.$100.\ ^{100}C_{49} - \ ^{100}C_{48}$
D.$100.\ ^{101}C_{49} - \ ^{100}C_{50}$
Solution
Let $1 + x = r$
$\therefore S = 1.r + 2.r^{2} + 3.r^{3} + \ldots\ldots + 100r^{100}\ldots\ldots$.
$$\begin{array}{r} rS = 1.r^{2} + 2.r^{3} + \ldots.. + 99r^{100} + 100r^{101}\#(AGP) \end{array}$$
(1) - (2) gives
$$S = - \frac{(1 + x)^{101}}{x^{2}} + \frac{1}{x^{2}} + \frac{100(1 + x)^{101}}{x} $$∴ coefficient $x^{48}$ in S
$= -$ coefficient of $x^{48}$ in $\frac{(1 + x)^{101}}{x^{2}} + 100$.
Coefficient of $x^{48}$ in $\frac{(1 + x)^{101}}{x}$
$$= 100\ ^{101}C_{49} - \ ^{101}C_{50}$$
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