Question 4 of 4beginner🔧 ApplyLong Answer4 marks

Verify that (1−x)−2=1+2x+3x2+4x3+…(1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \ldots (when ∣x∣<1|x| < 1).

Correct Answer

To verify (1−x)−2=1+2x+3x2+4x3+…(1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \ldots (when ∣x∣<1|x| < 1):

Step 1: Using the Binomial Theorem for (1+x)m(1+x)^m when mm is a negative integer, the expansion is:

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

Step 2: For (1−x)−2(1-x)^{-2}, substitute m=−2m = -2 and replace xx with −x-x:

(1−x)−2=1+(−2)(−x)+(−2)(−3)2!(−x)2+(−2)(−3)(−4)3!(−x)3+…(1-x)^{-2} = 1 + (-2)(-x) + \frac{(-2)(-3)}{2!}(-x)^2 + \frac{(-2)(-3)(-4)}{3!}(-x)^3 + \ldots

Step 3: Simplifying each term:

  • First term: 11
  • Second term: (−2)(−x)=2x(-2)(-x) = 2x
  • Third term: 62⋅x2=3x2\frac{6}{2} \cdot x^2 = 3x^2
  • Fourth term: −246⋅(−x3)=4x3\frac{-24}{6} \cdot (-x^3) = 4x^3

Step 4: Therefore, we get:

(1−x)−2=1+2x+3x2+4x3+…(1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \ldots

This expansion is valid when ∣x∣<1|x| < 1 (i.e., −1<x<1-1 < x < 1), as the condition ∣x∣<1|x| < 1 is necessary when mm is a negative integer. Hence verified.

Exercise: A.1.2 Binomial Theorem for any Index | Q: 4 | (Page Page 2)
For More Understanding

Explanation

The question requires applying the binomial expansion formula for negative integer exponents. From the context, the general expansion (a+b)m=am+mam−1b+m(m−1)1.2am−2b2+…(a+b)^m = a^m + ma^{m-1}b + \frac{m(m-1)}{1.2} a^{m-2}b^2 + \ldots applies.

The context explicitly states that the condition ∣x∣<1|x| < 1 is necessary when mm is a negative integer. By substituting m=−2m = -2 and carefully computing each coefficient, we verify the given expansion.

Solution Steps

  1. Step 1: Write the general binomial expansion for negative integer exponent

  2. Step 2: Substitute m = -2 and replace x with -x

  3. Step 3: Calculate each coefficient systematically

  4. Step 4: Verify the pattern matches the given series and state the validity condition

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