The coefficient of $x^{50}$ in the expression $(1 + x)^{100} + 2 x (1 + x)^{99} + 3 x^{2} (1 + x)^{98} + . . . . . . . . . + 101. x^{100}$ is:

^{102} C_{50}

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^{102} C_{51}

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^{105} C_{50}

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^{105} C_{49}

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Solution

The correct option is C $^{102} C_{50}$
The above expression is an Arithmetic geometric progression.
$S = (1 + x)^{100} + 2 x (1 + x)^{99} . . .101 x^{100}$ ...(i)
$x (1 + x)^{- 1} S = x (1 + x)^{99} + 2 x^{2} (1 + x)^{98} . . .101 x^{101} (1 + x)^{- 1}$ ...(ii)
Subtracting ii from i we get
$S (1 + x)^{- 1} = (1 + x)^{100} + x (1 + x)^{99} + x^{2} (1 + x)^{98} . . . x^{100} - 101 x^{101} (1 + x)^{- 1}$
$S = [(1 + x)^{101} + x (1 + x)^{100} + x^{2} (1 + x)^{99} . . . x^{100} (1 + x)] - 101 x^{101}$
The term in the square bracket is a G.P with common ration $\frac{x}{1 + x}$
Hence
$S = \frac{(1 + x)^{101} (1 - (\frac{x}{1 + x})^{101})}{1 - \frac{x}{1 + x}} - 101 x^{101}$
$= (1 + x)^{102} (1 - (\frac{x}{1 + x})^{101}) - 101 x^{101}$
$= (1 + x)^{102} - x^{101} (1 + x) - 101 x^{101}$
Hence coefficient of $x^{50}$ is $^{102} C_{50}$

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, s h o w t h a t

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