Question 1 of 20beginner🔧 ApplyLong Answer4 marks

Expand each of the expressions in Exercises 1 to 5.

(1−2x)5(1 - 2x)^5
Correct Answer

The expansion of the expression is 1−10x+40x2−80x3+80x4−32x5\mathbf{1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5}.

To expand the expression (1−2x)5(1 - 2x)^5, we use the binomial expansion for (x−y)n(x - y)^n provided in the context. By comparing (1−2x)5(1 - 2x)^5 with (x−y)n(x - y)^n, we identify x=1x = 1, y=2xy = 2x, and n=5n = 5.

Applying the formula:

(x−y)n=nC0 xn−nC1 xn−1y+nC2 xn−2y2−⋯+(−1)n nCn yn(x - y)^n = {}^nC_0\, x^n - {}^nC_1\, x^{n-1} y + {}^nC_2\, x^{n-2} y^2 - \cdots + (-1)^n\, {}^nC_n\, y^n

We substitute the values into the expansion:

(1−2x)5=5C0 (1)5−5C1 (1)4(2x)+5C2 (1)3(2x)2−5C3 (1)2(2x)3+5C4 (1)1(2x)4−5C5 (2x)5(1 - 2x)^5 = {}^5C_0\, (1)^5 - {}^5C_1\, (1)^4 (2x) + {}^5C_2\, (1)^3 (2x)^2 - {}^5C_3\, (1)^2 (2x)^3 + {}^5C_4\, (1)^1 (2x)^4 - {}^5C_5\, (2x)^5

Next, we calculate the binomial coefficients and simplify the terms:

  1. 5C0=1{}^5C_0 = 1: Term is 1(1)=11(1) = 1
  2. 5C1=5{}^5C_1 = 5: Term is −5(1)(2x)=−10x-5(1)(2x) = -10x
  3. 5C2=10{}^5C_2 = 10: Term is 10(1)(4x2)=40x210(1)(4x^2) = 40x^2
  4. 5C3=10{}^5C_3 = 10: Term is −10(1)(8x3)=−80x3-10(1)(8x^3) = -80x^3
  5. 5C4=5{}^5C_4 = 5: Term is 5(1)(16x4)=80x45(1)(16x^4) = 80x^4
  6. 5C5=1{}^5C_5 = 1: Term is −1(32x5)=−32x5-1(32x^5) = -32x^5

Combining all the terms, the final expansion is 1−10x+40x2−80x3+80x4−32x5\mathbf{1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5}.

Exercise: EXERCISE 7.1 | Q: 1 | (Page 7)
For More Understanding

Explanation

The question asks for the expansion of a binomial expression. The provided context includes the general formula for (x−y)n(x - y)^n and a specific worked example for (x−2y)5(x - 2y)^5. By mapping the problem (1−2x)5(1 - 2x)^5 to the formula where the first term is 11 and the second term is 2x2x, we can apply the binomial coefficients for n=5n = 5 (1,5,10,10,5,11, 5, 10, 10, 5, 1) to derive the solution step-by-step.

Solution Steps

  1. Step 1: Identify the values from the expression (1−2x)5(1 - 2x)^5 corresponding to the formula (x−y)n(x - y)^n. Here, the first term x=1x = 1, the second term y=2xy = 2x, and the power n=5n = 5.

  2. Step 2: Write the general expansion for (x−y)n(x - y)^n:

    nC0 xn−nC1 xn−1y+nC2 xn−2y2−⋯+(−1)n nCn yn{}^nC_0\, x^n - {}^nC_1\, x^{n-1} y + {}^nC_2\, x^{n-2} y^2 - \cdots + (-1)^n\, {}^nC_n\, y^n

  3. Step 3: Substitute the identified values into the expansion formula:

    5C0 (1)5−5C1 (1)4(2x)+5C2 (1)3(2x)2−5C3 (1)2(2x)3+5C4 (1)1(2x)4−5C5 (2x)5{}^5C_0\, (1)^5 - {}^5C_1\, (1)^4 (2x) + {}^5C_2\, (1)^3 (2x)^2 - {}^5C_3\, (1)^2 (2x)^3 + {}^5C_4\, (1)^1 (2x)^4 - {}^5C_5\, (2x)^5

  4. Step 4: Evaluate the binomial coefficients (5C0=1,  5C1=5,  5C2=10,  5C3=10,  5C4=5,  5C5=1)({}^5C_0 = 1,\; {}^5C_1 = 5,\; {}^5C_2 = 10,\; {}^5C_3 = 10,\; {}^5C_4 = 5,\; {}^5C_5 = 1) and simplify the powers of the terms.

  5. Step 5: Calculate the final product for each term:

    1−10x+40x2−80x3+80x4−32x51 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5
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