Expand Using the Binomial Theorem (x-1/2)^7

${(x - \frac{1}{2})}^{7}$

Step 1

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

$\sum_{k = 0}^{7} \frac{7!}{(7 - k)! k!} \cdot {(x)}^{7 - k} \cdot {(- \frac{1}{2})}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{7} \cdot {(- \frac{1}{2})}^{0} + 7 \cdot {(x)}^{6} \cdot {(- \frac{1}{2})}^{1} + \dots + 1 \cdot {(x)}^{0} \cdot {(- \frac{1}{2})}^{7}$

Step 4

Simplify the polynomial result.

$x^{7} - \frac{7 x^{6}}{2} + \frac{21 x^{5}}{4} - \frac{35 x^{4}}{8} + \frac{35 x^{3}}{16} - \frac{21 x^{2}}{32} + \frac{7 x}{64} - \frac{1}{128}$

Expand Using the Binomial Theorem (x-1/2)^7

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