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Q5:
Expand each of the expressions.
(x + \frac{1}{x})^{6}
Solution :
1. Theoretical Foundation: The Binomial Theorem
To expand the expression $(x + \frac{1}{x})^{6}$, we utilize the Binomial Theorem, which states that for any non-negative integer $n$:
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$
where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ represents the binomial coefficient [By the definition of Combinations].
2. Variable Assignment
For the given expression $(x + \frac{1}{x})^{6}$, we map the variables as follows:
- $a = x$
- $b = \frac{1}{x} = x^{-1}$
- $n = 6$
3. Step-by-Step Expansion
Applying the formula for $n=6$:
$(x + \frac{1}{x})^6 = \binom{6}{0}x^6(\frac{1}{x})^0 + \binom{6}{1}x^5(\frac{1}{x})^1 + \binom{6}{2}x^4(\frac{1}{x})^2 + \binom{6}{3}x^3(\frac{1}{x})^3 + \binom{6}{4}x^2(\frac{1}{x})^4 + \binom{6}{5}x^1(\frac{1}{x})^5 + \binom{6}{6}x^0(\frac{1}{x})^6$
4. Calculation of Coefficients and Simplification
| Term | Coefficient Calculation | Power Simplification | Result |
|---|---|---|---|
| 1st | $\binom{6}{0} = 1$ | $x^6 \cdot 1$ | $x^6$ |
| 2nd | $\binom{6}{1} = 6$ | $x^5 \cdot x^{-1} = x^4$ | $6x^4$ |
| 3rd | $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$ | $x^4 \cdot x^{-2} = x^2$ | $15x^2$ |
| 4th | $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ | $x^3 \cdot x^{-3} = 1$ | $20$ |
| 5th | $\binom{6}{4} = \binom{6}{2} = 15$ | $x^2 \cdot x^{-4} = x^{-2}$ | $15x^{-2}$ |
| 6th | $\binom{6}{5} = \binom{6}{1} = 6$ | $x^1 \cdot x^{-5} = x^{-4}$ | $6x^{-4}$ |
| 7th | $\binom{6}{6} = 1$ | $1 \cdot x^{-6}$ | $x^{-6}$ |
5. Visual Representation of Symmetry
Final Consolidation
Summing the components derived in the table:
$(x + \frac{1}{x})^6 = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$
Final Solution: x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}
More Questions from Class 11 Mathematics Binomial Theorem EXERCISE 7.1
- Q1: Expand each of the expressions 1. $(1-2x)^5$
- Q1: Expand each of the expressions. (1-2x)^{5}
- Q2: Expand each of the expressions 2. $(\frac{2}{x} - \frac{x}{2})^5$
- Q2: Expand each of the expressions. (\frac{2}{x} - \frac{x}{2})^{5}
- Q3: Expand each of the expressions $(2x - 3)^6$
- Q3: Expand each of the expressions. (2x - 3)^{6}
- Q4: Expand each of the expressions $(\frac{x}{3} + \frac{1}{x})^5$
- Q4: Expand each of the expressions. (\frac{x}{3} + \frac{1}{x})^{5}
- Q5: Expand each of the expressions 5. $(x + \frac{1}{x})^6$
- Q6: Using binomial theorem, evaluate each of the following: 6. $(96)^3$
- Q6: Using binomial theorem, evaluate each of the following: (96)^{3}
- Q7: Using binomial theorem, evaluate each of the following: 7. $(102)^5$
- Q7: Using binomial theorem, evaluate each of the following: (102)^{5}
- Q8: Using binomial theorem, evaluate each of the following: 8. $(101)^4$
- Q8: Using binomial theorem, evaluate each of the following: (101)^{4}
- Q9: Using binomial theorem, evaluate each of the following: 9. $(99)^5$
- Q9: Using binomial theorem, evaluate each of the following: (99)^{5}
- Q10: Using Binomial Theorem, indicate which number is larger $(1.1)^{10000}$ or 1000.
- Q10: Using Binomial Theorem, indicate which number is larger (1.1)^{10000} or 1000.
- Q11: Find $(a + b)^4 – (a – b)^4$. Hence, evaluate $(\sqrt{3} + \sqrt{2})^4 – (\sqrt{3} – \sqrt{2})^4$.
- Q11: Find ((a) + (b))^{4} – ((a) – (b))^{4}. Hence, evaluate ((\sqrt{3}+\sqrt{2})^{4} – (\sqrt{3}-\sqrt{2})^{4}).
- Q12: Find $(x + 1)^6 + (x – 1)^6$. Hence or otherwise evaluate $(\sqrt{2} + 1)^6 + (\sqrt{2} – 1)^6$.
- Q12: Find ((x) + 1)^{6} + ((x) – 1)^{6}. Hence or otherwise evaluate ((\sqrt{2} + 1)^{6} + (\sqrt{2} – 1)^{6}).
- Q13: Show that $9^{n+1} – 8n – 9$ is divisible by 64, whenever $n$ is a positive integer.
- Q13: Show that (9^{n+1} – 8n – 9) is divisible by 64, whenever (n) is a positive integer.
- Q14: Prove that $\sum_{r=0}^{n} {^nC_r} 3^r = 4^n$.
- Q14: Prove that (\sum_{r=0}^{n} {}^{n}C_{r} 3^{r} = 4^{n}).
CBSE Solutions for Class 11 Mathematics Binomial Theorem
Chapters in CBSE - Class 11 Mathematics
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