Question

Verify the first four terms of each infinite binomial series. $$(1-a)^{2 / 3}=1-2 a / 3-a^{2} / 9-4 a^{3} / 81 \ldots$$

Short answer

The first four terms are: \(1 - \frac{2}{3}a + \frac{1}{9}a^2 - \frac{4}{81}a^3\).

Step-by-step solution

- Recall the Binomial Theorem for non-integer exponents

The Binomial Theorem for any exponent states that \[ (1 - a)^n = 1 + \sum_{k=1}^\infty \frac{n(n-1)...(n-k+1)}{k!}(-a)^k .\]

- Apply the formula to the given power

Substitute the given exponent into the binomial theorem: \[ n = \frac{2}{3} \] and determine the first four terms. Use the general term formula: \[ \frac{n(n-1)...(n-k+1)}{k!}(-a)^k \] to calculate the terms.

- Calculate the first term

For \( k = 0 \), the first term is always \( 1 \) since anything to the power of zero is one.

- Calculate the second term

For \( k = 1 \), the second term is: \[ \frac{2/3}{1!}(-a)^1 = -\frac{2}{3}a .\]

- Calculate the third term

For \( k = 2 \), the third term is: \[ \frac{(2/3)(-1/3)}{2!}(-a)^2 = \frac{-2 \cdot -1}{3 \cdot 3 \cdot 2}a^2 = \frac{2}{18}a^2 = \frac{1}{9}a^2 .\]

- Calculate the fourth term

For \( k = 3 \), the fourth term is: \[ \frac{(2/3)(-1/3)(-4/3)}{3!}(-a)^3 = \frac{-2 \cdot -1 \cdot -4}{3 \cdot 3 \cdot 3 \cdot 6}a^3 = -\frac{8}{162}a^3 = -\frac{4}{81}a^3 .\]

Key concepts

Binomial Series Expansion

When we talk about binomial series expansion, we're referring to expressing a binomial expression like \( (1 - a)^n \) as a series. This isn't limited to positive integers; it can also involve fractions, irrational, or even negative numbers as exponents.

The general formula for a binomial expansion with non-integer exponents is given by:
\[\begin{equation}(1 - a)^n = 1 + \sum_{k=1}^\infty \frac{n(n-1)...(n-k+1)}{k!}(-a)^k\end{equation}\]
This infinite series continues indefinitely, with each term providing a more accurate approximation of the value of the binomial expression. To express the first few terms of a binomial series, we simply calculate each term based on its position in the series. For the exercise at hand, the terms are derived using the aforementioned formula, taking care to manage the signs and arithmetic involving the non-integer exponent \(\frac{2}{3}\).

Mathematical Induction

Though not directly employed in this exercise, mathematical induction is a proof technique that can be very useful when working with series, including binomial series expansion. It involves proving that a statement holds for a successive chain of numbers, starting with a base case and then proving that if the case holds for one number \(n\), it must also hold for \(n + 1\).

For instance, if we wanted to prove the validity of our binomial expansion for all terms of the series, we could use induction to show that each term follows a predictable pattern based on our general term formula:
\[\begin{equation}\frac{n(n-1)...(n-k+1)}{k!}(-a)^k\end{equation}\]
By proving that the series expansion correctly calculates one term, and then proves that the next term follows suit, induction helps to establish the correctness of the entire series, even though we often work with only the first few terms.

Infinite Series

An infinite series is a sum of infinitely many terms, and it is crucial in the field of calculus and higher mathematics. The binomial series expansion results in an infinite series especially when the exponent is not an integer.

The sequence of partial sums of an infinite series may approach a limit, which is the series' value if it converges. For example, the binomial series:
\[\begin{equation}(1 - a)^{n} = 1 + \sum_{k=1}^{\infty} \frac{n(n-1)...(n-k+1)}{k!}(-a)^{k}\end{equation}\]
represents an infinite series that converges to the value \( (1 - a)^n \) as long as \( |a| < 1 \) for real numbers. This concept underpins much of analysis and is crucial for understanding the behavior of functions within calculus. By familiarizing oneself with infinite series, a student can gain insight into the approximation of functions and the sum of their infinite terms.