Question 12 of 20intermediate🔧 ApplyLong Answer4 marks

Expand each of the expressions in Exercises 1 to 5.

(x3+1x)5\left(\frac{x}{3} + \frac{1}{x}\right)^5

Correct Answer

Using the Binomial Theorem,

(a+b)n=nC0 an+nC1 an−1b+nC2 an−2b2+…+nCn bn.(a + b)^n = {}^nC_0\, a^n + {}^nC_1\, a^{n-1}b + {}^nC_2\, a^{n-2}b^2 + \ldots + {}^nC_n\, b^n.

For the expression (x3+1x)5\left(\frac{x}{3} + \frac{1}{x}\right)^5, we have a=x3a = \frac{x}{3}, b=1xb = \frac{1}{x}, and n=5n = 5.

Step 1: Write the expansion using binomial coefficients.

(x3+1x)5=5C0(x3)5+5C1(x3)4(1x)+5C2(x3)3(1x)2+5C3(x3)2(1x)3+5C4(x3)(1x)4+5C5(1x)5\left(\frac{x}{3} + \frac{1}{x}\right)^5 = {}^5C_0 \left(\frac{x}{3}\right)^5 + {}^5C_1 \left(\frac{x}{3}\right)^4 \left(\frac{1}{x}\right) + {}^5C_2 \left(\frac{x}{3}\right)^3 \left(\frac{1}{x}\right)^2 + {}^5C_3 \left(\frac{x}{3}\right)^2 \left(\frac{1}{x}\right)^3 + {}^5C_4 \left(\frac{x}{3}\right) \left(\frac{1}{x}\right)^4 + {}^5C_5 \left(\frac{1}{x}\right)^5

Step 2: Substitute the values of binomial coefficients: 5C0=1{}^5C_0 = 1, 5C1=5{}^5C_1 = 5, 5C2=10{}^5C_2 = 10, 5C3=10{}^5C_3 = 10, 5C4=5{}^5C_4 = 5, 5C5=1{}^5C_5 = 1.

Step 3: Simplify each term.

  • First term: 1⋅x5243=x52431 \cdot \frac{x^5}{243} = \frac{x^5}{243}
  • Second term: 5⋅x481⋅1x=5x3815 \cdot \frac{x^4}{81} \cdot \frac{1}{x} = \frac{5x^3}{81}
  • Third term: 10⋅x327⋅1x2=10x2710 \cdot \frac{x^3}{27} \cdot \frac{1}{x^2} = \frac{10x}{27}
  • Fourth term: 10⋅x29⋅1x3=109x10 \cdot \frac{x^2}{9} \cdot \frac{1}{x^3} = \frac{10}{9x}
  • Fifth term: 5⋅x3⋅1x4=53x35 \cdot \frac{x}{3} \cdot \frac{1}{x^4} = \frac{5}{3x^3}
  • Sixth term: 1⋅1x5=1x51 \cdot \frac{1}{x^5} = \frac{1}{x^5}

Final Answer:

(x3+1x)5=x5243+5x381+10x27+109x+53x3+1x5\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}

Exercise: EXERCISE 7.1 | Q: 4 | (Chapter: 8)
For More Understanding

Explanation

This question is from Exercise 7.1, Question 4. The solution applies the standard binomial expansion formula

(a+b)n=∑r=0nnCr an−rbr(a+b)^n = \sum_{r=0}^{n} {}^nC_r\, a^{n-r}b^r

as shown in the context. The key steps involve identifying a=x/3a = x/3 and b=1/xb = 1/x, then systematically expanding and simplifying each term while maintaining the sum of indices of aa and bb equal to n=5n=5 in each term.

Solution Steps

  1. Step 1: Identify a=x3a = \frac{x}{3}, b=1xb = \frac{1}{x}, n=5n = 5 and write the binomial expansion formula.

  2. Step 2: Substitute binomial coefficients: 5C0=1{}^5C_0=1, 5C1=5{}^5C_1=5, 5C2=10{}^5C_2=10, 5C3=10{}^5C_3=10, 5C4=5{}^5C_4=5, 5C5=1{}^5C_5=1.

  3. Step 3: Expand each term by applying powers to numerator and denominator separately.

  4. Step 4: Simplify each term by combining like powers of xx.

  5. Step 5: Write the final expansion with all six terms in descending powers of xx.

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