Q.28
Let nCr denote the number n!r!(n−r)!. Then for n=100, the sum of the series
1−nC1+nC2−nC3+⋯+(−1)rnCr+⋯+(−1)nnCn
is
(A) 0
(B) 1
(C) 2
(D) 1024
Quick Answer
The given series is an alternating binomial sum for \(n=100\).
Its value is 0, so the correct option is (A) 0.
Problem Statement
The problem defines \(^{n}C_r = \frac{n!}{r!(n-r)!}\) and asks, for \(n=100\), the value of the series:
\[1 – ^{100}C_1 + ^{100}C_2 – ^{100}C_3 + \cdots + (-1)^r ^{100}C_r + \cdots + (-1)^{100} ^{100}C_{100}\]
with options:
- (A) 0
- (B) 1
- (C) 2
- (D) 1024
Core Concept: Binomial Theorem
Use the binomial theorem for \((1+x)^n\):
\[(1+x)^n = \sum_{r=0}^n ^{n}C_r x^r\]
Now substitute \(x = -1\):
\[(1-1)^n = \sum_{r=0}^n ^{n}C_r (-1)^r\]
Left side: \((1-1)^n = 0^n = 0\) for \(n \geq 1\).
Right side is exactly the given series: \(^{n}C_0 – ^{n}C_1 + ^{n}C_2 – \cdots + (-1)^n ^{n}C_n\).
Therefore: \[^{100}C_0 – ^{100}C_1 + ^{100}C_2 – \cdots + ^{100}C_{100} = 0\]
Explanation of Each Option
Option (A) 0
Matches the exact value obtained from the binomial theorem for \((1-1)^{100} = 0\).
This is the correct answer.
Option (B) 1
A value of 1 would occur for sums like \(\sum_{r=0}^n ^{n}C_r = 2^n\) or for expressions such as \((1)^n\).
Here the alternating signs correspond to \((1-1)^n\), which is 0, not 1.
Option (C) 2
The number 2 is often associated with \((1+1)^1 = 2\) or as a factor of \(2^n\).
The alternating binomial expansion does not simplify to 2 for \(n=100\).
Option (D) 1024
Note that \(1024 = 2^{10}\), commonly seen in binomial problems with \(n=10\).
Our problem has \(n=100\) and an alternating series; the value is 0, not a power of 2.
Brief Introduction
The alternating binomial coefficient series for n = 100 is a classic problem that tests conceptual understanding of the binomial theorem and alternating sums. By recognizing the series as the expansion of \((1-1)^{100}\), one can quickly show that the entire expression evaluates to zero, while also understanding why tempting distractors like 1, 2, and 1024 are wrong.
