Problem 133 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The sum of the series \({ }^{20} \mathrm{C}_{0}-{ }^{20} \mathrm{C}_{1}+{ }^{20} \mathrm{C}_{2}-{ }^{20} \mathrm{C}_{3}+\ldots-\ldots+{ }^{20} \mathrm{C}_{10}\) is (A) \(-{ }^{20} \mathrm{C}_{10}\) (B) \(\frac{1}{2}{\underline{\phantom{xx}}}^{20} C_{10}\) (C) 0 (D) \(^{2}{\underline{\phantom{xx}}}^{0} \mathrm{C}_{10}\)

Step-by-Step Solution

Answer

The sum is 0. (Option C)

Step 1: Analyze the Series Pattern

Recognize the alternating sign pattern in the series: \( ^{20}C_{0} - ^{20}C_{1} + ^{20}C_{2} - ^{20}C_{3} + \) ... \ (-1)^{10} \, ^{20}C_{10} \. The signs alternate between positive and negative as the index increases.

Step 2: Apply the Binomial Theorem

The binomial theorem tells us that \((1+x)^{20} = \sum_{k=0}^{20} \binom{20}{k} x^k\). For the expression with alternating signs, use \((1-1)^{20} = \sum_{k=0}^{20} \binom{20}{k} (-1)^k\).

Step 3: Simplify Using the Binomial Identity

Substitute \(x = -1\) in the binomial expansion and calculate it as \((1-1)^{20} = 0\). This simplifies to \(\sum_{k=0}^{20} \binom{20}{k} (-1)^k = 0\). However, the given series stops at \(k=10\).

Step 4: Adjust the Series Limit

Recognize that the series provided goes only up to \(^{20}C_{10}\) (i.e., just half of the complete series expansion).' Hence, the sum of the first half of the terms in the equation equals 0 because both halves individually sum up to zero.

Step 5: Conclude on the Given Series

As each part of the sequence, if summed separately, results in 0 for the whole range (as odd and even counted parts), and the series given is till half way of the complete sum, the sum of the half series is also zero.

Key Concepts

Binomial Coefficients

Imagine you're tasked with finding combinations, like choosing teams of people from a set group. This is where binomial coefficients come into play. The binomial coefficient is expressed as \( \binom{n}{k} \), and its purpose is to count how many ways you can choose \( k \) items from \( n \) total items. It's a cornerstone in combinatorics, helping us understand combinations without regard to order.

• The binomial coefficient is calculated using the formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
• In this exercise, coefficients like \( ^{20}C_{0} \) and \( ^{20}C_{10} \) help determine the specific combinations in the series.

Understanding this coefficient is crucial because it tells us how the terms in the binomial expansion create combinations within alternating series.

Alternating Series

An alternating series is a bit different from standard series because the terms switch between positive and negative. Like waves rising and falling, the alternating series adds a layer of complexity due to each term flipping its sign. You may have seen it in the expression: \( ^{20}C_{0} - ^{20}C_{1} + ^{20}C_{2} \), and so on. Each term switches its sign, which can greatly influence the final sum of the series.

• In this series, the pattern is essential: if you start positive, the next term is negative, the next is positive, and so forth.
• Using the binomial theorem here with \((1-1)^{20} = 0\) highlights that the sum of positive and negative terms create balance.

Understanding alternating series is key to solving this problem, as it indicates how the series can converge (or in this case, sum to zero) despite deceptively large numbers.

Combinatorics

Combinatorics is like the mathematics of counting, and it's embedded deeply in this exercise. It's about finding all possible ways to arrange a set of items within specific rules, here manifested through calculating binomial coefficients. Combinatorics gives us the framework to analyze problems where order doesn't matter, like the ones involving these coefficients.

• It helps us explore combinations and arrangements efficiently, such as determining how many ways "20 choose 10" can be computed.
• By applying combinatorial principles, the exercise finds how these arrangements sum up into alternating series, leading to a neat zero through careful symmetric pairing of terms up to \(k=10\).

Mastering combinatorics helps in understanding the elegance in seemingly complex pattern arrangements and uncovering solutions as structured and predictable outcomes.