Expand Using the Binomial Theorem (2+5n^2)^2

${(2 + 5 n^{2})}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(2)}^{2 - k} \cdot {(5 n^{2})}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(2)}^{2 - 0} \cdot {(5 n^{2})}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(2)}^{2 - 1} \cdot {(5 n^{2})}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(2)}^{2 - 2} \cdot {(5 n^{2})}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2)}^{2} \cdot {(5 n^{2})}^{0} + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Step 4

Simplify each term.

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

${(2)}^{2} \cdot {(5 n^{2})}^{0} + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

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

$4 \cdot {(5 n^{2})}^{0} + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Apply the product rule to $5 n^{2}$ .

$4 \cdot (5^{0} {(n^{2})}^{0}) + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Anything raised to $0$ is $1$ .

$4 \cdot (1 {(n^{2})}^{0}) + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Multiply ${(n^{2})}^{0}$ by $1$ .

$4 \cdot {(n^{2})}^{0} + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Multiply the exponents in ${(n^{2})}^{0}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 \cdot n^{2 \cdot 0} + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Multiply $2$ by $0$ .

$4 \cdot n^{0} + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Anything raised to $0$ is $1$ .

$4 \cdot 1 + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Multiply $4$ by $1$ .

$4 + 2 \cdot {(2)}^{1} \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Evaluate the exponent.

$4 + 2 \cdot 2 \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Multiply $2$ by $2$ .

$4 + 4 \cdot {(5 n^{2})}^{1} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Simplify.

$4 + 4 \cdot (5 n^{2}) + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Multiply $5$ by $4$ .

$4 + 20 n^{2} + 1 \cdot {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Multiply ${(2)}^{0}$ by $1$ .

$4 + 20 n^{2} + {(2)}^{0} \cdot {(5 n^{2})}^{2}$

Anything raised to $0$ is $1$ .

$4 + 20 n^{2} + 1 \cdot {(5 n^{2})}^{2}$

Multiply ${(5 n^{2})}^{2}$ by $1$ .

$4 + 20 n^{2} + {(5 n^{2})}^{2}$

Apply the product rule to $5 n^{2}$ .

$4 + 20 n^{2} + 5^{2} {(n^{2})}^{2}$

Raise $5$ to the power of $2$ .

$4 + 20 n^{2} + 25 {(n^{2})}^{2}$

Multiply the exponents in ${(n^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 + 20 n^{2} + 25 n^{2 \cdot 2}$

Multiply $2$ by $2$ .

$4 + 20 n^{2} + 25 n^{4}$

Expand Using the Binomial Theorem (2+5n^2)^2

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