Question 1 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 or fractional indices.

According to the theorem:

(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

This expansion holds whenever ∣x∣<1|x| < 1.

For (1+x)−1(1+x)^{-1}, we substitute m=−1m = -1:

First term: 11

Second term: mx=(−1)x=−xmx = (-1)x = -x

Third term:

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

Fourth term:

m(m−1)(m−2)1.2.3x3=(−1)(−2)(−3)6x3=−x3\frac{m(m-1)(m-2)}{1.2.3} x^3 = \frac{(-1)(-2)(-3)}{6} x^3 = -x^3

Continuing this pattern, each successive term alternates in sign with increasing powers of xx.

Therefore, we obtain:

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

The condition ∣x∣<1|x| < 1 is necessary because when the index is negative, there are infinite number of terms in the expansion, and the series converges only when ∣x∣<1|x| < 1.

Exercise: A.1.2 Binomial Theorem for any Index | Q: 1 | (Page Page 2)
For More Understanding

Explanation

The question asks to verify a specific case of the Binomial Theorem with negative index. The textbook provides the general formula for (1+x)m(1+x)^m when mm is negative or fractional, and explicitly lists (1+x)−1=1−x+x2−x3+…(1+x)^{-1} = 1 - x + x^2 - x^3 + \ldots as a particular case.

The verification involves substituting m=−1m = -1 into the general formula and computing 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/fractional indices:

    (1+x)m=1+mx+m(m−1)2x2+m(m−1)(m−2)6x3+…(1+x)^m = 1 + mx + \frac{m(m-1)}{2}x^2 + \frac{m(m-1)(m-2)}{6}x^3 + \ldots

  2. Step 2: Substitute m=−1m = -1 into the formula

  3. Step 3: Calculate each term: 1st term = 11, 2nd term = −x-x, 3rd term = x2x^2, 4th term = −x3-x^3

  4. Step 4: Observe the alternating sign pattern and write the complete expansion

  5. Step 5: State the condition ∣x∣<1|x| < 1 required for convergence

Next