Question 15 of 20advanced🔧 ApplyLong Answer5 marks

If aa and bb are distinct integers, prove that a−ba - b is a factor of an−bna^n - b^n, whenever nn is a positive integer.

Correct Answer

To Prove: If aa and bb are distinct integers, then (a−b)(a - b) is a factor of an−bna^n - b^n, where nn is a positive integer.

Proof:

We are given that aa and bb are distinct integers and nn is a positive integer. We need to prove that (a−b)(a - b) divides an−bna^n - b^n.

Following the hint, we write an=(a−b+b)na^n = (a - b + b)^n and expand using the Binomial Theorem.

By the Binomial Theorem:

(a−b+b)n=nC0⋅(a−b)n+nC1⋅(a−b)n−1⋅b+nC2⋅(a−b)n−2⋅b2+⋯+nCn⋅bn(a - b + b)^n = {}^nC_0 \cdot (a-b)^n + {}^nC_1 \cdot (a-b)^{n-1} \cdot b + {}^nC_2 \cdot (a-b)^{n-2} \cdot b^2 + \cdots + {}^nC_n \cdot b^n

Therefore:

an=(a−b)n+n⋅(a−b)n−1⋅b+nC2⋅(a−b)n−2⋅b2+⋯+bna^n = (a-b)^n + n \cdot (a-b)^{n-1} \cdot b + {}^nC_2 \cdot (a-b)^{n-2} \cdot b^2 + \cdots + b^n

Now, subtracting bnb^n from both sides:

an−bn=(a−b)n+n⋅(a−b)n−1⋅b+nC2⋅(a−b)n−2⋅b2+⋯+nCn−1⋅(a−b)⋅bn−1a^n - b^n = (a-b)^n + n \cdot (a-b)^{n-1} \cdot b + {}^nC_2 \cdot (a-b)^{n-2} \cdot b^2 + \cdots + {}^nC_{n-1} \cdot (a-b) \cdot b^{n-1}

Each term on the right-hand side contains (a−b)(a-b) as a factor. Factoring out (a−b)(a-b):

an−bn=(a−b)[(a−b)n−1+n⋅(a−b)n−2⋅b+nC2⋅(a−b)n−3⋅b2+⋯+nCn−1⋅bn−1]a^n - b^n = (a-b)\left[(a-b)^{n-1} + n \cdot (a-b)^{n-2} \cdot b + {}^nC_2 \cdot (a-b)^{n-3} \cdot b^2 + \cdots + {}^nC_{n-1} \cdot b^{n-1}\right]

Since the expression in brackets is an integer (as aa, bb are integers and binomial coefficients are integers), we conclude that (a−b)(a - b) is a factor of an−bna^n - b^n.

∴ Hence proved.

Exercise: Miscellaneous Exercise on Chapter 7 | Q: 1 | (Page 8)
For More Understanding

Explanation

This question is from the Miscellaneous Exercise on Chapter 7 (Binomial Theorem). The textbook provides the hint to write an=(a−b+b)na^n = (a - b + b)^n and expand. The key insight is that by expanding (a−b+b)n(a-b+b)^n using the Binomial Theorem, every term except the last one (bnb^n) contains (a−b)(a-b) as a factor. When we subtract bnb^n from both sides, all remaining terms have (a−b)(a-b) as a common factor, which can be factored out to prove the statement.

Solution Steps

  1. Step 1: Write an=(a−b+b)na^n = (a - b + b)^n using the given hint.

  2. Step 2: Expand (a−b+b)n(a - b + b)^n using the Binomial Theorem formula.

  3. Step 3: Subtract bnb^n from both sides to get an−bna^n - b^n.

  4. Step 4: Observe that each term contains (a−b)(a-b) as a factor.

  5. Step 5: Factor out (a−b)(a-b) to show it divides an−bna^n - b^n.

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