Question 2 of 4beginner🔧 ApplyLong Answer4 marks

Verify that (1−x)−1=1+x+x2+x3+…(1-x)^{-1} = 1 + x + x^2 + x^3 + \ldots (when ∣x∣<1|x| < 1).

Correct Answer

To verify (1−x)−1=1+x+x2+x3+…(1-x)^{-1} = 1 + x + x^2 + x^3 + \ldots when ∣x∣<1|x| < 1, we use the Binomial Theorem for negative indices.

The theorem states that for ∣x∣<1|x| < 1:

(1+x)m=1+mx+m(m−1)1.2x2+m(m−1)(m−2)1.2.3x3+…(1 + x)^m = 1 + mx + \frac{m(m-1)}{1.2} x^2 + \frac{m(m-1)(m-2)}{1.2.3} x^3 + \ldots

For (1−x)−1(1-x)^{-1}, we substitute m=−1m = -1 and replace xx by (−x)(-x):

(1−x)−1=(1+(−x))−1(1-x)^{-1} = (1 + (-x))^{-1}

Calculating each term:

  • First term: 11
  • Second term: (−1)(−x)=x(-1)(-x) = x
  • Third term: (−1)(−2)1.2⋅(−x)2=22⋅x2=x2\frac{(-1)(-2)}{1.2} \cdot (-x)^2 = \frac{2}{2} \cdot x^2 = x^2
  • Fourth term: (−1)(−2)(−3)1.2.3⋅(−x)3=−66⋅(−x3)=x3\frac{(-1)(-2)(-3)}{1.2.3} \cdot (-x)^3 = \frac{-6}{6} \cdot (-x^3) = x^3

Continuing this pattern, we obtain:

(1−x)−1=1+x+x2+x3+…(1-x)^{-1} = 1 + x + x^2 + x^3 + \ldots

This expansion is valid when ∣x∣<1|x| < 1. Hence verified.

Exercise: A.1.2 Binomial Theorem for any Index | Q: 2 | (Chapter: Page 2)
For More Understanding

Explanation

The question asks to verify the expansion of (1−x)−1(1-x)^{-1} using the Binomial Theorem. The textbook context provides the general formula for (1+x)m(1+x)^m when mm is negative or fractional, and also lists this specific expansion as a particular case.

Students need to substitute m=−1m = -1 and replace xx with −x-x in the general formula, then calculate the coefficients of each term systematically. The condition ∣x∣<1|x| < 1 is essential for convergence of the infinite series.

Solution Steps

  1. Step 1: Write the Binomial Theorem formula for negative indices:

    (1+x)m=1+mx+m(m−1)1.2x2+…(1 + x)^m = 1 + mx + \frac{m(m-1)}{1.2} x^2 + \ldots

  2. Step 2: Substitute m=−1m = -1 and replace xx by (−x)(-x) to get (1−x)−1(1-x)^{-1}

  3. Step 3: Calculate each coefficient: first term = 11, second term = (−1)(−x)=x(-1)(-x) = x, third term = (−1)(−2)2⋅x2=x2\frac{(-1)(-2)}{2} \cdot x^2 = x^2, fourth term = (−1)(−2)(−3)6⋅(−x3)=x3\frac{(-1)(-2)(-3)}{6} \cdot (-x^3) = x^3

  4. Step 4: Observe the pattern continues as 1+x+x2+x3+…1 + x + x^2 + x^3 + \ldots

  5. Step 5: State that the expansion is valid when ∣x∣<1|x| < 1

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