Problem 66 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\left({ }^{(\mathrm{m}} C_{0}+{ }^{m} C_{1}-{ }^{m} C_{2}-{ }^{m} C_{3}\right)+\left({ }^{\mathrm{m}} C_{4}+{ }^{m} C_{5}-{ }^{m} C_{6}-{ }^{m} C_{7}\right)+\ldots\) \(=0\) if and only if for some positive integer \(k, m=\) (A) \(4 k\) (B) \(4 k+1\) (C) \(4 k-1\) (D) \(4 k+2\)

Step-by-Step Solution

Answer

(A) 4k

Step 1: Understanding the Pattern

The expression is a series of terms that cycle through combinations of an integer, expressed as \(^m C_n\). The first cycle is \(^m C_0\), \(^m C_1\), \(^m C_2\), \(^m C_3\). The second cycle is \(^m C_4\), \(^m C_5\), \(^m C_6\), \(^m C_7\), and so forth. Each cycle consists of two positive and two negative terms.

Step 2: Analyzing a Single Block

The terms in each cycle are combined such that their sum is \(({}^{m}C_0 + {}^{m}C_1 - {}^{m}C_2 - {}^{m}C_3)\), \(({}^{m}C_4 + {}^{m}C_5 - {}^{m}C_6 - {}^{m}C_7)\), etc. When we sum a cycle \(({}^{m}C_{4k} + {}^{m}C_{4k+1} - {}^{m}C_{4k+2} - {}^{m}C_{4k+3}) = 0\), it implies symmetrical balancing based on the binomial theorem.

Step 3: Applying the Binomial Theorem

According to the binomial theorem, the sum \(({}^{m}C_0 + {}^{m}C_1 + {}^{m}C_2 + ... + {}^{m}C_m)\) equals \(2^m\). The expression rearranges these terms to evaluate to zero. The sum of alternating terms equates to half the sequence when \(m\) is divisible by four, as this causes equal distribution.

Step 4: Identifying the Condition for Zero Sum

The total expression equals zero when terms are grouped symmetrically across the binomial coefficients. This pattern, called the Pascal's Ladder, insists on the sum being zero only if the factors allow equal positive and negative distribution among component groups in alternating cycles.

Step 5: Concluding Condition (m divisible by 4)

For the entire series of terms to add up to zero, \(m\) must be such that the sum of each block resolves symmetrically. This condition occurs when \(m\) is a multiple of 4, i.e., \(m = 4k\) for some positive integer \(k\).

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool in mathematics used to expand expressions that are raised to a power. It's particularly useful when dealing with expressions of the form \[(a + b)^n.\]When expanded, this expression results in a series of terms involving binomial coefficients. This theorem allows us to understand how coefficients are distributed in polynomial expansions. In simple terms, each term in the expansion is determined by the formula \[^{n}C_{k} \, a^{n-k} \, b^{k},\]where \(^nC_k\) represents the binomial coefficient.
These coefficients show up as part of a pattern related to the powers of \(a\) and \(b\). Importantly, they hint at symmetrical properties within binomial expansions. When we study problems like the given exercise, the symmetry property of these coefficients underlies why we can potentially rearrange the sequence and achieve a zero sum under certain conditions.

Pascal's Triangle

Pascal's Triangle is a simple yet profound mathematical triangle of numbers. Each number is the sum of the two numbers directly above it in the previous row. This triangle is not just a neat numerical curiosity, but it also efficiently represents binomial coefficients, which underscore many algebraic expansions and combinatorial calculations.

• The first row of Pascal's Triangle has a single number: 1
• The second row has two 1's
• Every subsequent row begins and ends with 1, and each interior element is the sum of two numbers from the row above

These rows correspond to the coefficients in the expansion of a binomial expression. For example, row 3 follows from expanding \[(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3.\] The triangle reflects the balance and symmetry in the arrangement of terms. It also underpins many combinatorial identities used in analyzing and simplifying expressions in algebra.

Binomial Coefficients

Binomial coefficients are central in combinatorics, as they give the number of ways to choose \(k\) elements from a set of \(n\) elements, often denoted as \(^nC_k\). These coefficients can be expressed using the formula:\[^{n}C_{k} = \frac{n!}{k!(n-k)!},\]where \(!\) denotes a factorial, the product of all positive integers up to that number.
In the context of algebra, binomial coefficients play a critical role. They are used to determine terms in the expansions of binomials, as laid out by the Binomial Theorem. When exploring patterns within these coefficients, especially through the lens of problems like in our exercise, one sees the potential for symmetry and cancellation when coefficients are summed in structured ways.
Understanding the equivalent distributions of binomial coefficients in alternating cycles of sums, such as \((^mC_0 + ^mC_1 - ^mC_2 - ^mC_3),\) as described originally, demonstrates intricate balance points facilitated by these coefficients across mathematical expressions and sequences.