Expand Using Pascal's Triangle (2z+5k)^4

${(2 z + 5 k)}^{4}$

Step 1

Pascal\'s Triangle can be displayed as such:

$1$

$1 - 1$

$1 - 2 - 1$

$1 - 3 - 3 - 1$

$1 - 4 - 6 - 4 - 1$

The triangle can be used to calculate the coefficients of the expansion of ${(a + b)}^{n}$ by taking the exponent $n$ and adding $1$ . The coefficients will correspond with line $n + 1$ of the triangle. For ${(2 z + 5 k)}^{4}$ , $n = 4$ so the coefficients of the expansion will correspond with line $5$ .

Step 2

The expansion follows the rule ${(a + b)}^{n} = c_{0} a^{n} b^{0} + c_{1} a^{n - 1} b^{1} + c_{n - 1} a^{1} b^{n - 1} + c_{n} a^{0} b^{n}$ . The values of the coefficients, from the triangle, are $1 - 4 - 6 - 4 - 1$ .

$1 a^{4} b^{0} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + 1 a^{0} b^{4}$

Step 3

Substitute the actual values of $a$ $2 z$ and $b$ $5 k$ into the expression.

$1 {(2 z)}^{4} {(5 k)}^{0} + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Step 4

Simplify each term.

Multiply ${(2 z)}^{4}$ by $1$ .

${(2 z)}^{4} {(5 k)}^{0} + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Apply the product rule to $2 z$ .

$2^{4} z^{4} {(5 k)}^{0} + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Raise $2$ to the power of $4$ .

$16 z^{4} {(5 k)}^{0} + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Apply the product rule to $5 k$ .

$16 z^{4} (5^{0} k^{0}) + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 \cdot 5^{0} z^{4} k^{0} + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 \cdot 1 z^{4} k^{0} + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Multiply $16$ by $1$ .

$16 z^{4} k^{0} + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 z^{4} \cdot 1 + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Multiply $16$ by $1$ .

$16 z^{4} + 4 {(2 z)}^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Apply the product rule to $2 z$ .

$16 z^{4} + 4 (2^{3} z^{3}) {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Raise $2$ to the power of $3$ .

$16 z^{4} + 4 (8 z^{3}) {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Multiply $8$ by $4$ .

$16 z^{4} + 32 z^{3} {(5 k)}^{1} + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Simplify.

$16 z^{4} + 32 z^{3} (5 k) + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 z^{4} + 32 \cdot 5 z^{3} k + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Multiply $32$ by $5$ .

$16 z^{4} + 160 z^{3} k + 6 {(2 z)}^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Apply the product rule to $2 z$ .

$16 z^{4} + 160 z^{3} k + 6 (2^{2} z^{2}) {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

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

$16 z^{4} + 160 z^{3} k + 6 (4 z^{2}) {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Multiply $4$ by $6$ .

$16 z^{4} + 160 z^{3} k + 24 z^{2} {(5 k)}^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Apply the product rule to $5 k$ .

$16 z^{4} + 160 z^{3} k + 24 z^{2} (5^{2} k^{2}) + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 z^{4} + 160 z^{3} k + 24 \cdot 5^{2} z^{2} k^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

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

$16 z^{4} + 160 z^{3} k + 24 \cdot 25 z^{2} k^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Multiply $24$ by $25$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 4 {(2 z)}^{1} {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Simplify.

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 4 (2 z) {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Multiply $2$ by $4$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 8 z {(5 k)}^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Apply the product rule to $5 k$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 8 z (5^{3} k^{3}) + 1 {(2 z)}^{0} {(5 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 8 \cdot 5^{3} z k^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Raise $5$ to the power of $3$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 8 \cdot 125 z k^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

Multiply $8$ by $125$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 1 {(2 z)}^{0} {(5 k)}^{4}$

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

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + {(2 z)}^{0} {(5 k)}^{4}$

Apply the product rule to $2 z$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 2^{0} z^{0} {(5 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 1 z^{0} {(5 k)}^{4}$

Multiply $z^{0}$ by $1$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + z^{0} {(5 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 1 {(5 k)}^{4}$

Multiply ${(5 k)}^{4}$ by $1$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + {(5 k)}^{4}$

Apply the product rule to $5 k$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 5^{4} k^{4}$

Raise $5$ to the power of $4$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 625 k^{4}$

Expand Using Pascal's Triangle (2z+5k)^4

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