Question

For the expression \((1-x)^{100}\). Then sum of coefficient of first 50 terms is:

• A. \(^{99}C_{49}\)
• B. \(-\frac{^{100}C_{50}}{2}\)
• C. \(-^{99}C_{49}\)
• D. \(-^{101}C_{50}\)

Hint:

thebinomial theoremandrelatedidentitiesforsimplifyingsumsof binomialcoefficients. Thesumofthefirstntermsinabinomialexpansioncan oftenbeexpressedinaclosedformusingcombinations

Answer 1

The Correct Option is B

Expand the binomial expression: \[ (1 - x)^{100} = C_0 - C_1x + C_2x^2 - C_3x^3 + \cdots - C_{99}x^{99} + C_{100}x^{100}. \]

Collecting the coefficients: \[ C_0 - C_1 + C_2 - C_3 + \cdots - C_{99} + C_{100} = 0. \]

Now consider twice the coefficients: \[ 2(C_0 - C_1 + C_2 - \cdots - C_9) + C_{50} = 0. \]

Simplify further: \[ C_0 - C_1 + C_2 - \cdots + C_{99} = -\frac{1}{2} C_{50}. \]

Using the properties of binomial coefficients: \[ C_{50} = \binom{100}{50}, \quad C_0 - C_1 + C_2 - \cdots + C_{99} = -\frac{1}{2} \binom{100}{50}. \]

Finally, calculate: \[ C_{49} = \frac{1}{2} \frac{100!}{50! \cdot 50!} = -\frac{99}{49}. \]

Thus: \[ C_0 - C_1 + C_2 - \cdots + C_{99} = -\frac{99}{49}. \]

Answer 2

Sum of coefficient of first 50 terms

\((t)\) \(= ^{100}C_0-^{100}C_1+...+^{100}C_{40}\)

Now

\(^{100}C_0-^{100}C_1+...+^{100}C_{100}=0\)

\(2[^{100}C_0-^{100}C_1+...]+ ^{100}C_{50}=0\)

\(\therefore \; t= -\frac{1}{2}^{100}C_{50}\)

The correct option is (B): \(-\frac{^{100}C_{50}}{2}\)