Expand Using Pascal's Triangle (s+t)^4

${(s + t)}^{4}$

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 ${(s + t)}^{4}$ , $n = 4$ 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 $a$ $s$ and $b$ $t$ into the expression.

$1 {(s)}^{4} {(t)}^{0} + 4 {(s)}^{3} {(t)}^{1} + 6 {(s)}^{2} {(t)}^{2} + 4 {(s)}^{1} {(t)}^{3} + 1 {(s)}^{0} {(t)}^{4}$

Step 4

Simplify each term.

Multiply ${(s)}^{4}$ by $1$ .

${(s)}^{4} {(t)}^{0} + 4 {(s)}^{3} {(t)}^{1} + 6 {(s)}^{2} {(t)}^{2} + 4 {(s)}^{1} {(t)}^{3} + 1 {(s)}^{0} {(t)}^{4}$

Anything raised to $0$ is $1$ .

$s^{4} \cdot 1 + 4 {(s)}^{3} {(t)}^{1} + 6 {(s)}^{2} {(t)}^{2} + 4 {(s)}^{1} {(t)}^{3} + 1 {(s)}^{0} {(t)}^{4}$

Multiply $s^{4}$ by $1$ .

$s^{4} + 4 {(s)}^{3} {(t)}^{1} + 6 {(s)}^{2} {(t)}^{2} + 4 {(s)}^{1} {(t)}^{3} + 1 {(s)}^{0} {(t)}^{4}$

Simplify.

$s^{4} + 4 s^{3} t + 6 {(s)}^{2} {(t)}^{2} + 4 {(s)}^{1} {(t)}^{3} + 1 {(s)}^{0} {(t)}^{4}$

Simplify.

$s^{4} + 4 s^{3} t + 6 s^{2} t^{2} + 4 s {(t)}^{3} + 1 {(s)}^{0} {(t)}^{4}$

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

$s^{4} + 4 s^{3} t + 6 s^{2} t^{2} + 4 s t^{3} + {(s)}^{0} {(t)}^{4}$

Anything raised to $0$ is $1$ .

$s^{4} + 4 s^{3} t + 6 s^{2} t^{2} + 4 s t^{3} + 1 {(t)}^{4}$

Multiply ${(t)}^{4}$ by $1$ .

$s^{4} + 4 s^{3} t + 6 s^{2} t^{2} + 4 s t^{3} + t^{4}$

Expand Using Pascal's Triangle (s+t)^4

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