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Expand (1 - 1/2x)6 to the fourth term.

Mathematics

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Answer

1−3x+154x2−52x31 - 3x + \frac{15}{4}x^2 - \frac{5}{2}x^3

Step-by-step solution

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To expand (1−12x)6(1 - \frac{1}{2}x)^6 to the fourth term, we use the binomial theorem, which states that:

(a+b)n=(n0)anb0+(n1)an−1b1+(n2)an−2b2+(n3)an−3b3+…(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \binom{n}{3}a^{n-3}b^3 + \dots

In this case, a=1a=1, b=−12xb=-\frac{1}{2}x, and n=6n=6.

  1. Calculate the first term (k=0k=0).

    Term 1=(60)(1)6(−12x)0\text{Term 1} = \binom{6}{0}(1)^6\left(-\frac{1}{2}x\right)^0 Term 1=1×1×1=1\text{Term 1} = 1 \times 1 \times 1 = 1

  2. Calculate the second term (k=1k=1).

    Term 2=(61)(1)5(−12x)1\text{Term 2} = \binom{6}{1}(1)^5\left(-\frac{1}{2}x\right)^1 Term 2=6×1×(−12x)=−3x\text{Term 2} = 6 \times 1 \times \left(-\frac{1}{2}x\right) = -3x

  3. Calculate the third term (k=2k=2).

    Term 3=(62)(1)4(−12x)2\text{Term 3} = \binom{6}{2}(1)^4\left(-\frac{1}{2}x\right)^2 (62)=6×52×1=15\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 Term 3=15×1×(14x2)=154x2\text{Term 3} = 15 \times 1 \times \left(\frac{1}{4}x^2\right) = \frac{15}{4}x^2

  4. Calculate the fourth term (k=3k=3).

    Term 4=(63)(1)3(−12x)3\text{Term 4} = \binom{6}{3}(1)^3\left(-\frac{1}{2}x\right)^3 (63)=6×5×43×2×1=20\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 Term 4=20×1×(−18x3)=−208x3=−52x3\text{Term 4} = 20 \times 1 \times \left(-\frac{1}{8}x^3\right) = -\frac{20}{8}x^3 = -\frac{5}{2}x^3

  5. Combine the terms.

    The expansion to the fourth term is the sum of these terms:

    1−3x+154x2−52x31 - 3x + \frac{15}{4}x^2 - \frac{5}{2}x^3

    The expansion of (1−12x)6(1 - \frac{1}{2}x)^6 to the fourth term is 1−3x+154x2−52x3\boxed{1 - 3x + \frac{15}{4}x^2 - \frac{5}{2}x^3}.

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