The sum of the coefficients of the first 50 terms in the binomial expansion of ( 1 - x ) 100 , is equal to [2023]

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Solution

Q.
The sum of the coefficients of the first 50 terms in the binomial expansion of ${(1 - x)}^{100},$ is equal to [2023]

1 ${}^{101}{C_{50}}$

2 $- {}^{99}{C_{49}}$

3 $- {}^{101}{C_{50}}$

4 ${}^{99}{C_{49}}$

Ans.
(2)

${(1 - x)}^{100} = {}^{100}{C_{0}} - {}^{100}{C_{1}} x + {}^{100}{C_{2}} x^{2} - {}^{100}{C_{3}} x^{3} + \dots + {}^{100}{C_{100}} x^{100}$

or
${(1 - x)}^{100} = C_{0} - C_{1} x + C_{2} x^{2} - C_{3} x^{3} + \dots + C_{100} x^{100}$

If $x$ = 1, then

$0 = C_{0} - C_{1} + C_{2} - C_{3} + \dots - C_{99} + C_{100}$

$\Rightarrow 0 = 2 (C_{0} - C_{1} + C_{2} - \dots - C_{49}) + C_{50}$

$\Rightarrow C_{0} - C_{1} + C_{2} - \dots - C_{49} = \frac{- C_{50}}{2} = \frac{- 1}{2} \frac{100!}{50! 50!} = \frac{- 99!}{50! 49!}$

$\Rightarrow C_{0} - C_{1} + C_{2} - \dots - C_{49} = - {}^{99}{C_{49}}$

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