Expand Using the Binomial Theorem (1-x/2)^8

${(1 - \frac{x}{2})}^{8}$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{8} \frac{8!}{(8 - k)! k!} \cdot {(1)}^{8 - k} \cdot {(- \frac{x}{2})}^{k}$

Step 2

Expand the summation.

$\frac{8!}{(8 - 0)! 0!} \cdot {(1)}^{8 - 0} \cdot {(- \frac{x}{2})}^{0} + \frac{8!}{(8 - 1)! 1!} \cdot {(1)}^{8 - 1} \cdot {(- \frac{x}{2})}^{1} + \dots + \frac{8!}{(8 - 8)! 8!} \cdot {(1)}^{8 - 8} \cdot {(- \frac{x}{2})}^{8}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(1)}^{8} \cdot {(- \frac{x}{2})}^{0} + 8 \cdot {(1)}^{7} \cdot {(- \frac{x}{2})}^{1} + \dots + 1 \cdot {(1)}^{0} \cdot {(- \frac{x}{2})}^{8}$

Step 4

Simplify the polynomial result.

$1 - 4 x + 7 x^{2} - 7 x^{3} + \frac{35 x^{4}}{8} - \frac{7 x^{5}}{4} + \frac{7 x^{6}}{16} - \frac{x^{7}}{16} + \frac{x^{8}}{256}$

Expand Using the Binomial Theorem (1-x/2)^8

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