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Q4:
Expand each of the expressions $(\frac{x}{3} + \frac{1}{x})^5$
Expand each of the expressions $(\frac{x}{3} + \frac{1}{x})^5$
Solution :
Step 1: Application of the Binomial Theorem
For any positive integer $n$, the Binomial Theorem states that: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ [Per the Binomial Theorem identity].
Given the expression $\left(\frac{x}{3} + \frac{1}{x}\right)^5$, we identify the variables as follows:
- $a = \frac{x}{3}$
- $b = \frac{1}{x}$
- $n = 5$
Step 2: Expansion of Terms
Substituting the values into the general formula: $\left(\frac{x}{3} + \frac{1}{x}\right)^5 = \binom{5}{0} \left(\frac{x}{3}\right)^5 + \binom{5}{1} \left(\frac{x}{3}\right)^4 \left(\frac{1}{x}\right)^1 + \binom{5}{2} \left(\frac{x}{3}\right)^3 \left(\frac{1}{x}\right)^2 + \binom{5}{3} \left(\frac{x}{3}\right)^2 \left(\frac{1}{x}\right)^3 + \binom{5}{4} \left(\frac{x}{3}\right)^1 \left(\frac{1}{x}\right)^4 + \binom{5}{5} \left(\frac{1}{x}\right)^5$
Step 3: Calculating Binomial Coefficients and Powers
We use the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ [Definition of Combinations].
| Term | Coefficient | Variable Part |
|---|---|---|
| $k=0$ | $\binom{5}{0} = 1$ | $\frac{x^5}{243}$ |
| $k=1$ | $\binom{5}{1} = 5$ | $5 \cdot \frac{x^4}{81} \cdot \frac{1}{x} = \frac{5x^3}{81}$ |
| $k=2$ | $\binom{5}{2} = 10$ | $10 \cdot \frac{x^3}{27} \cdot \frac{1}{x^2} = \frac{10x}{27}$ |
| $k=3$ | $\binom{5}{3} = 10$ | $10 \cdot \frac{x^2}{9} \cdot \frac{1}{x^3} = \frac{10}{9x}$ |
| $k=4$ | $\binom{5}{4} = 5$ | $5 \cdot \frac{x}{3} \cdot \frac{1}{x^4} = \frac{5}{3x^3}$ |
| $k=5$ | $\binom{5}{5} = 1$ | $\frac{1}{x^5}$ |
Step 4: Algebraic Simplification
Aggregating the simplified terms computed in Step 3:
$\left(\frac{x}{3} + \frac{1}{x}\right)^5 = \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$
Final Solution: \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}
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}
- Q5: Expand each of the expressions 5. $(x + \frac{1}{x})^6$
- Q5: Expand each of the expressions. (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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