How to Simplify a Binomial Expansion using Differentiation

How to Simplify Binomial Expansion using Differentiation

Navigating the intricacies of binomial expansions in mathematics can sometimes feel like treading through a complex labyrinth. However, fear not, as we are about to embark on a journey that will unravel the secrets of simplifying binomial expansions with the powerful tool of differentiation. Binomial expansions are fundamental in various fields of mathematics and science, and the ability to simplify them can significantly streamline problem-solving processes. In this guide, we will delve into the step-by-step process of simplifying binomial expansions using differentiation, empowering you to tackle mathematical challenges with confidence and precision. Whether you’re a student looking to ace your exams or a math enthusiast seeking a deeper understanding, this tutorial will equip you with valuable skills to make binomial expansions more manageable and intuitive. So, let’s embark on this mathematical adventure and discover the art of simplifying binomial expansions through differentiation.

Simplifying a binomial expansion using differentiation is a calculus technique to find a more concise form of a two-term equation. This method, often referred to as ‘binomial expansion by differentiation,’ involves taking the derivative of the expanded expression. It’s a powerful tool for making complex expressions more manageable and is particularly useful in mathematical analysis and problem-solving.

Example

Show that \( \displaystyle \begin{pmatrix} n \\ 2 \end{pmatrix} 2^2 + \begin{pmatrix} n \\ 4 \end{pmatrix} 4^2 + \begin{pmatrix} n \\ 6 \end{pmatrix} 6^2 + \cdots + \begin{pmatrix} n \\ n \end{pmatrix} n^2 = n(n+1) 2^{n-3} \), for \( n \ge 4 \) and assuming that \( n \) is even.

\( \displaystyle \begin{align} \begin{pmatrix} n \\ 0 \end{pmatrix} + \begin{pmatrix} n \\ 1 \end{pmatrix} x + \begin{pmatrix} n \\ 2 \end{pmatrix} x^2 + \begin{pmatrix} n \\ 3 \end{pmatrix} x^3 + \begin{pmatrix} n \\ 4 \end{pmatrix} x^4 + \cdots + \begin{pmatrix} n \\ n \end{pmatrix} x^n &= (1+x)^{n} \\ &\text{binomial expansion} \\ \begin{pmatrix} n \\ 1 \end{pmatrix} + \begin{pmatrix} n \\ 2 \end{pmatrix} 2x + \begin{pmatrix} n \\ 3 \end{pmatrix} 3x^2 + \begin{pmatrix} n \\ 4 \end{pmatrix} 4x^3 + \cdots + \begin{pmatrix} n \\ n \end{pmatrix} nx^{n-1} &= n(1+x)^{n-1} \\ &\text{differentiate} \\ \begin{pmatrix} n \\ 1 \end{pmatrix} x + \begin{pmatrix} n \\ 2 \end{pmatrix} 2x^2 + \begin{pmatrix} n \\ 3 \end{pmatrix} 3x^3 + \begin{pmatrix} n \\ 4 \end{pmatrix} 4x^4 + \cdots + \begin{pmatrix} n \\ n \end{pmatrix} nx^{n} &= nx(1+x)^{n-1} \\ &\text{multiply } x \text{ both sides} \\ \begin{pmatrix} n \\ 1 \end{pmatrix} + \begin{pmatrix} n \\ 2 \end{pmatrix} 2^2x + \begin{pmatrix} n \\ 3 \end{pmatrix} 3^2x^2 + \begin{pmatrix} n \\ 4 \end{pmatrix} 4^2x^3 + \cdots + \begin{pmatrix} n \\ n \end{pmatrix} n^2x^{n-1} &= n(n-1)x(1+x)^{n-2}+n(1+x)^{n-1} \\ &\text{differentiate} \\ &= n(1+x)^{n-2}\left[ x(n-1)+(1+x) \right] \\ &= n(1+x)^{n-2}(nx-x+1+x) \\ &= n(nx+1)(1+x)^{n-2} \cdots (1) \\ \begin{pmatrix} n \\ 1 \end{pmatrix} + \begin{pmatrix} n \\ 2 \end{pmatrix} 2^2 + \begin{pmatrix} n \\ 3 \end{pmatrix} 3^2 + \begin{pmatrix} n \\ 4 \end{pmatrix} 4^2 + \cdots + \begin{pmatrix} n \\ n \end{pmatrix} n^2 &= n(n+1)2^{n-2} \cdots (2) \\ &x=1 \text{ into } (1) \\ \begin{pmatrix} n \\ 1 \end{pmatrix}-\begin{pmatrix} n \\ 2 \end{pmatrix} 2^2 + \begin{pmatrix} n \\ 3 \end{pmatrix} 3^2-\begin{pmatrix} n \\ 4 \end{pmatrix} 4^2 + \cdots + (-1)^{n-1} \begin{pmatrix} n \\ n \end{pmatrix} n^2 &= n(-n+1)(1-1)^{n-2} \\ &x=-1 \text{ into } (1) \\ \begin{pmatrix} n \\ 1 \end{pmatrix}-\begin{pmatrix} n \\ 2 \end{pmatrix} 2^2 + \begin{pmatrix} n \\ 3 \end{pmatrix} 3^2-\begin{pmatrix} n \\ 4 \end{pmatrix} 4^2 + \cdots-\begin{pmatrix} n \\ n \end{pmatrix} n^2 &= 0 \cdots (3) \\ &n-1 \text{ is odd, thus } (-1)^{n-1}=-1 \\ 2 \begin{pmatrix} n \\ 2 \end{pmatrix}2^2 + 2\begin{pmatrix} n \\ 4 \end{pmatrix} 4^2 + 2\begin{pmatrix} n \\ 6 \end{pmatrix} 6^2 + \cdots + 2\begin{pmatrix} n \\ n \end{pmatrix} n^2 &= n(n+1)2^{n-2} \\ \therefore \begin{pmatrix} n \\ 2 \end{pmatrix}2^2 + \begin{pmatrix} n \\ 4 \end{pmatrix} 4^2 + \begin{pmatrix} n \\ 6 \end{pmatrix} 6^2 + \cdots + \begin{pmatrix} n \\ n \end{pmatrix} n^2 &= n(n+1)2^{n-3} \end{align} \)

 

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