The coefficient of $x^{50}$ in the binomial expansion of $(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + . . . + x^{1000}$ is :

\frac{(1000)!}{(49)! (951)!}

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\frac{(1001)!}{(51)! (950)!}

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\frac{(1000)!}{(50)! (950)!}

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\frac{(1001)!}{(50)! (951)!}

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Solution

The correct option is D $\frac{(1001)!}{(50)! (951)!}$
$S = (1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + . . . + x^{1000}$

This is a Geometric Series.

$a = (1 + x)^{1000}$

$r = \frac{x}{x + 1}$

$n = 1001$

$S = \frac{((1 + x)^{1000}) (1 - \frac{x^{1001}}{(x + 1)^{1001}})}{1 - \frac{x}{1 + x}} = (1 + x)^{1001} - x^{1001}$

Coefficient of $x^{50}$ is $^{1001} C_{50} = \frac{(1001)!}{(951)! (50)!}$ .

Suggest Corrections

Similar questions

x^{50}

(1 + x)^{1000} + 2 x (1 + x)^{999} + 3 x^{2} (1 + x)^{998} + . . . . . . + 1001 x^{1000}

x^{50}

(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + . . . + x^{1000}

x^{50}

(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + . . . + x^{1000}

Simplify the expression
${(1 + x)}^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots + x^{1000}$ and find the coefficient of $x^{50}$

x^{50}

(x > 0)

(1 + x)^{1000} + 2 x (1 + x)^{999} + 3 x^{2} (1 + x)^{998} + \dots

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