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

To find the sum of the coefficients of the first 50 terms of the expansion of \((1-x)^{100}\), we need to analyze the binomial expansion. The binomial expansion of \((1-x)^{100}\) is given by:

\((1-x)^{100} = \sum_{k=0}^{100} (-1)^k \cdot ^{100}C_k \cdot x^k\)

We are interested in finding the sum of the coefficients of the first 50 terms, which means we consider terms from \(k=0\) to \(k=49\).

The sum of the coefficients of these first 50 terms is:

\(S = ^{100}C_0 - ^{100}C_1 + ^{100}C_2 - ... + (-1)^{49} \cdot ^{100}C_{49}\)

This can be simplified using the property of binomial coefficients:

\(\sum_{k=0}^{49} (-1)^k \cdot ^{100}C_k = \frac{1}{2} \left( \sum_{k=0}^{100} (-1)^k \cdot ^{100}C_k \right)\)

According to the binomial theorem:

\(\sum_{k=0}^{100} (-1)^k \cdot ^{100}C_k = (1 - 1)^{100} = 0\)

This implies:

\(\sum_{k=0}^{100} (-1)^k \cdot ^{100}C_k = 0\)

Thus, we have:

\(\sum_{k=0}^{49} (-1)^k \cdot ^{100}C_k = -\sum_{k=50}^{100} (-1)^k \cdot ^{100}C_k\)

Using \(\sum_{k=0}^{100} (-1)^k \cdot ^{100}C_k = 0\), and splitting the sum:

\(\sum_{k=50}^{100} (-1)^k \cdot ^{100}C_k = -\sum_{k=0}^{49} (-1)^k \cdot ^{100}C_k\)

Therefore, we need to compute half of the complete sum when \(k\) ranges from 0 to 100:

\(\sum_{k=0}^{49} (-1)^k \cdot ^{100}C_k = -\frac{1}{2} \sum_{k=50}^{100} (-1)^k \cdot ^{100}C_k\)

The complete solution gives:

\(S = -\frac{^{100}C_{50}}{2}\)

The correct answer is \(- \frac{^{100}C_{50}}{2}\), which matches option B.