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Q1:
If (a) and (b) are distinct integers, prove that (a – b) is a factor of (a^{n} – b^{n}), whenever (n) is a positive integer.
[Hint write (a^{n} = (a – b + b)^{n}) and expand]
Solution :
Step 1: Theoretical Foundation via Binomial Expansion
To prove that $(a - b)$ is a factor of $(a^n - b^n)$, we utilize the Binomial Theorem, which states that for any positive integer $n$:
$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$
where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ [Per the Binomial Theorem Definition].
Step 2: Substitution and Expansion
Following the hint, we express $a$ as $(a - b + b)$. Thus, we consider the expression $a^n$ rewritten as:
$a^n = ((a - b) + b)^n$
Applying the Binomial Theorem by setting $x = (a - b)$ and $y = b$:
$a^n = \sum_{k=0}^{n} \binom{n}{k} (a - b)^{n-k} b^k$
$a^n = \binom{n}{0}(a - b)^n b^0 + \binom{n}{1}(a - b)^{n-1} b^1 + \dots + \binom{n}{n-1}(a - b)^1 b^{n-1} + \binom{n}{n}(a - b)^0 b^n$
Step 3: Algebraic Simplification
We observe that all terms in the summation contain at least one factor of $(a - b)$, except for the final term where $(a - b)^0 = 1$:
$a^n = [(a - b)^n + \binom{n}{1}(a - b)^{n-1} b + \dots + \binom{n}{n-1}(a - b) b^{n-1}] + b^n$
Factor out $(a - b)$ from all terms within the brackets:
$a^n = (a - b) \left[ (a - b)^{n-1} + \binom{n}{1}(a - b)^{n-2} b + \dots + \binom{n}{n-1} b^{n-1} \right] + b^n$
Step 4: Final Derivation of the Factor
Subtract $b^n$ from both sides of the equation:
$a^n - b^n = (a - b) \left[ \sum_{k=0}^{n-1} \binom{n}{k} (a - b)^{n-k-1} b^k \right]$
Since the expression within the square brackets is a sum of products of integers $(a, b, n)$, it results in an integer, let us denote it as $I$. Therefore:
$a^n - b^n = (a - b) \cdot I$
By the definition of divisibility, if $a^n - b^n$ can be expressed as the product of $(a - b)$ and an integer $I$, then $(a - b)$ must be a factor of $(a^n - b^n)$ [Per the Euclidean Property of Integers].
Visual Representation of Factorization Logic
Final Solution: Since $(a^n - b^n) = (a - b) \times I$, it is proven that $(a - b)$ is a factor of $(a^n - b^n)$ for all positive integers $n$.
More Questions from Class 11 Mathematics Binomial Theorem Miscellaneous Exercise on Chapter 7
- Q2: Evaluate ((\sqrt{3} + \sqrt{2})^{6} – (\sqrt{3} – \sqrt{2})^{6}).
- Q3: Find the value of ((a^{2} + \sqrt{a^{2}-1})^{4} + (a^{2} - \sqrt{a^{2}-1})^{4}).
- Q4: Find an approximation of ((0.99)^{5}) using the first three terms of its expansion.
- Q5: Expand using Binomial Theorem (1 + \frac{x}{2} - \frac{2}{x})^{4}, (x \neq 0).
- Q6: Find the expansion of ((3x^{2} – 2ax + 3a^{2})^{3}) using binomial theorem.
CBSE Solutions for Class 11 Mathematics Binomial Theorem
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