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Q2:
Evaluate ((\sqrt{3} + \sqrt{2})^{6} – (\sqrt{3} – \sqrt{2})^{6}).
Solution :
Step 1: Application of the Binomial Theorem
According to the Binomial Theorem, the expansion of $(a+b)^n$ is given by:
$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$Let $x = \sqrt{3}$ and $y = \sqrt{2}$. The given expression is of the form $(x+y)^6 - (x-y)^6$.
The expansion for $(x+y)^6$ is: $(x+y)^6 = \binom{6}{0}x^6 + \binom{6}{1}x^5y + \binom{6}{2}x^4y^2 + \binom{6}{3}x^3y^3 + \binom{6}{4}x^2y^4 + \binom{6}{5}xy^5 + \binom{6}{6}y^6$
The expansion for $(x-y)^6$ is: $(x-y)^6 = \binom{6}{0}x^6 - \binom{6}{1}x^5y + \binom{6}{2}x^4y^2 - \binom{6}{3}x^3y^3 + \binom{6}{4}x^2y^4 - \binom{6}{5}xy^5 + \binom{6}{6}y^6$
Step 2: Simplifying the Difference
Subtracting $(x-y)^6$ from $(x+y)^6$, the even-positioned terms (relative to the index $k$ starting at 0) cancel out, leaving twice the sum of the odd-indexed terms:
$(x+y)^6 - (x-y)^6 = 2 \left[ \binom{6}{1}x^5y + \binom{6}{3}x^3y^3 + \binom{6}{5}xy^5 \right]$ [By the property of binomial symmetry where $(a+b)^n - (a-b)^n = 2 \sum_{k \text{ odd}} \binom{n}{k} a^{n-k} b^k$]Step 3: Calculating Binomial Coefficients
| Combination | Calculation | Result |
|---|---|---|
| $\binom{6}{1}$ | $\frac{6!}{1!5!}$ | $6$ |
| $\binom{6}{3}$ | $\frac{6 \times 5 \times 4}{3 \times 2 \times 1}$ | $20$ |
| $\binom{6}{5}$ | $\binom{6}{1}$ | $6$ |
Step 4: Substitution and Evaluation
Substitute $x = \sqrt{3}$ and $y = \sqrt{2}$ into the simplified expression:
$= 2 \left[ 6(\sqrt{3})^5(\sqrt{2}) + 20(\sqrt{3})^3(\sqrt{2})^3 + 6(\sqrt{3})(\sqrt{2})^5 \right]$Evaluate the powers:
- $(\sqrt{3})^5 = 9\sqrt{3}$; $(\sqrt{2}) = \sqrt{2} \implies 6(9\sqrt{3})(\sqrt{2}) = 54\sqrt{6}$
- $(\sqrt{3})^3 = 3\sqrt{3}$; $(\sqrt{2})^3 = 2\sqrt{2} \implies 20(3\sqrt{3})(2\sqrt{2}) = 120\sqrt{6}$
- $(\sqrt{3}) = \sqrt{3}$; $(\sqrt{2})^5 = 4\sqrt{2} \implies 6(\sqrt{3})(4\sqrt{2}) = 24\sqrt{6}$
Step 5: Visual Representation
Final Solution: 396\sqrt{6}
More Questions from Class 11 Mathematics Binomial Theorem Miscellaneous Exercise on Chapter 7
- 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]
- 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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