Problem 18 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The sum of the last ten coefficients in the expansion of \((1+x)^{19}\) when expanded in ascending powers of \(x\) is (A) \(2^{18}\) (B) \(2^{19}\) (C) \(2^{18}-{ }^{19} C_{10}\) (D) \(\frac{1}{2}\left(2^{19}-1\right)\)

Step-by-Step Solution

Answer

(C) \(2^{18} - ^{19}C_{10}\)

Step 1: Identify the Problem

We need to find the sum of the last ten coefficients in the expansion of the binomial term \((1+x)^{19}\).

Step 2: Understanding the Coefficients

The binomial expansion \((1+x)^{19}\) is given by the terms \(^{19}C_0, ^{19}C_1, ..., ^{19}C_{19}\). We are interested in the coefficients from \(^{19}C_{10}\) to \(^{19}C_{19}\).

Step 3: Apply the Property of Symmetrical Coefficients

The coefficients are symmetrical, so \(^{19}C_{k} = ^{19}C_{19-k}\). Thus, the sum of the last ten coefficients is the same as the sum of the first ten coefficients: \(^1^{9}C_0 + ^{19}C_1 + ... + ^{19}C_9\).

Step 4: Use Binomial Theorem Property

By the binomial theorem, \((1+1)^{19} = 2^{19}\) which equals the sum of all coefficients. We know the full sum equals \(2^{19}\).

Step 5: Calculate the Required Sum

Since the sum of all coefficients, \(2^{19}\), is split symmetrically, the sum of the first ten coefficients equals the sum of the last ten. Therefore, each half is \(\frac{1}{2} \times 2^{19} = 2^{18}\).

Step 6: Consider the Middle Coefficient Adjustment

Since there is one overlap term \(^{19}C_{9}\) in the middle pair combining twice, subtract it out to find the true half-sum. Thus, the sum is \(2^{18} - ^{19}C_{10}\).

Key Concepts

Binomial Coefficients

Binomial coefficients are the numerical factors that are used in the expansion of a binomial raised to a power. They are denoted as \( ^nC_r \), which reads as "n choose r". This represents the number of ways to choose \( r \) elements from a set of \( n \) elements, without considering the order. To visualize this, think of a pascal's triangle, where each number is the sum of the two directly above it. Binomial coefficients can be found in the rows, with each row corresponding to an expansion of the binomial theorem. For instance, the row corresponding to \((1 + x)^{19}\) is filled with coefficients \(^19C_0, ^{19}C_1,..., ^{19}C_{19}\). These coefficients have a crucial role in problems involving expansions and combinatorics, helping us find exact values without having to perform bulky calculations.

Symmetrical Property

The symmetrical property of binomial coefficients is a unique and fascinating characteristic. It tells us that the coefficients on either side of the binomial expansion are the same. Mathematically, this is expressed as \(^nC_k = ^nC_{n-k}\). This symmetry means that if you know one coefficient, you automatically know its counterpart on the other side. In the context of the exercise, when dealing with expansions like \((1 + x)^{19}\), it implies that the sum of the coefficients from \(^19C_0\) to \(^19C_9\) is identical to the sum of those from \(^19C_{10}\) to \(^19C_{19}\). Such properties become handy when calculating simplified solutions by reducing what you need to compute, as seen in this exercise where only the first ten coefficients need to be summed, thanks to their symmetrical nature.

Combinatorics

Combinatorics is the study of counting, arranging, and finding patterns. It's essential in various fields, from computer science to probability theory. Within combinatorics, binomial expansions and coefficients serve as fundamental tools. In solving the given problem, combinatorial principles help determine the number of ways to choose coefficients from a binomial expansion. Here, we utilized the concept of symmetry and split the complete sum of all coefficients using \[(1 + x)^n = \sum_{k=0}^{n} {^nC_k} x^k \].This tells us that for any expansion, the sum of coefficients is equal to the expansion evaluated at \(x = 1\). Thus, for \((1 + x)^{19}\), we have \(2^{19}\), which we divide in half (owing to symmetry) to find the sum of either half of the coefficients.By understanding these combinatorial strategies, otherwise complex arithmetic involving binomial theorem becomes simpler and more computationally friendly.