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

${(x + \frac{1}{2})}^{2}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{2} \cdot {(\frac{1}{2})}^{0} + 2 \cdot {(x)}^{1} \cdot {(\frac{1}{2})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Step 4

Simplify each term.

Multiply ${(x)}^{2}$ by $1$ .

${(x)}^{2} \cdot {(\frac{1}{2})}^{0} + 2 \cdot {(x)}^{1} \cdot {(\frac{1}{2})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Apply the product rule to $\frac{1}{2}$ .

$x^{2} \cdot \frac{1^{0}}{2^{0}} + 2 \cdot {(x)}^{1} \cdot {(\frac{1}{2})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Anything raised to $0$ is $1$ .

$x^{2} \cdot \frac{1}{2^{0}} + 2 \cdot {(x)}^{1} \cdot {(\frac{1}{2})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Anything raised to $0$ is $1$ .

$x^{2} \cdot \frac{1}{1} + 2 \cdot {(x)}^{1} \cdot {(\frac{1}{2})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Divide $1$ by $1$ .

$x^{2} \cdot 1 + 2 \cdot {(x)}^{1} \cdot {(\frac{1}{2})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Multiply $x^{2}$ by $1$ .

$x^{2} + 2 \cdot {(x)}^{1} \cdot {(\frac{1}{2})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Simplify.

$x^{2} + 2 \cdot x \cdot {(\frac{1}{2})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Simplify.

$x^{2} + 2 x \cdot \frac{1}{2} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

$x^{2} + 2 (x) \cdot \frac{1}{2} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Cancel the common factor.

$x^{2} + 2 x \cdot \frac{1}{2} + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Rewrite the expression.

$x^{2} + x + 1 \cdot {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Multiply ${(x)}^{0}$ by $1$ .

$x^{2} + x + {(x)}^{0} \cdot {(\frac{1}{2})}^{2}$

Anything raised to $0$ is $1$ .

$x^{2} + x + 1 \cdot {(\frac{1}{2})}^{2}$

Multiply ${(\frac{1}{2})}^{2}$ by $1$ .

$x^{2} + x + {(\frac{1}{2})}^{2}$

Apply the product rule to $\frac{1}{2}$ .

$x^{2} + x + \frac{1^{2}}{2^{2}}$

One to any power is one.

$x^{2} + x + \frac{1}{2^{2}}$

Raise $2$ to the power of $2$ .

$x^{2} + x + \frac{1}{4}$

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

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