7.3 Q-3

Exercise 7.3 — Question 3

This question involves expanding expressions using the Binomial Theorem for positive integer powers and simplifying the result.

Binomial Theorem (Positive Integer $n$)

For any positive integer $n$:

$(a+b{)}^n={\sum }_{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})a^{n-r}{b}^r$

where the binomial coefficient is:

$\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$

Binomial Series (Fractional or Negative $n$)

When $n$ is a fraction or negative integer, the expansion becomes an infinite series:

$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+\cdots$

Convergence condition: This series is valid only when $|x|<1$.

Expanding $(a+x{)}^n$ for General $a$

To apply the standard series formula, factor out $a^n$:

$(a+x{)}^n=a^n{\left(1+\frac{x}{a}\right)}^n$

The series then applies to ${\left(1+\frac{x}{a}\right)}^n$ provided $\left|\frac{x}{a}\right|<1$.

Worked Example

Expand $(3-2x{)}^{-2}$ as a binomial series and state the range of validity.

Step 1: Factor out $3^{-2}$:

$(3-2x{)}^{-2}=3^{-2}{\left(1-\frac{2x}{3}\right)}^{-2}=\frac{1}{9}{\left(1-\frac{2x}{3}\right)}^{-2}$

Step 2: Apply the binomial series with $n=-2$ and $u=-\frac{2x}{3}$:

${\left(1+u\right)}^{-2}=1+(-2)u+\frac{(-2)(-3)}{2!}{u}^2+\frac{(-2)(-3)(-4)}{3!}{u}^3+\cdots$

$=1-2u+3u^2-4u^3+\cdots$

Step 3: Substitute $u=-\frac{2x}{3}$:

$=1+\frac{4x}{3}+\frac{16x^2}{3}+\cdots$

Step 4: Multiply by $\frac{1}{9}$:

$(3-2x{)}^{-2}=\frac{1}{9}\left(1+\frac{4x}{3}+\frac{16x^2}{3}+\cdots \right)$

Validity: $\left|\frac{2x}{3}\right|<1\Rightarrow |x|<\frac{3}{2}$