Expand Using the Binomial Theorem (a+b)^12

${(a + b)}^{12}$

Step 1

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

$\sum_{k = 0}^{12} \frac{12!}{(12 - k)! k!} \cdot {(a)}^{12 - k} \cdot {(b)}^{k}$

Step 2

Expand the summation.

$\frac{12!}{(12 - 0)! 0!} {(a)}^{12 - 0} \cdot {(b)}^{0} + \frac{12!}{(12 - 1)! 1!} {(a)}^{12 - 1} \cdot {(b)}^{1} + \dots + \frac{12!}{(12 - 12)! 12!} {(a)}^{12 - 12} \cdot {(b)}^{12}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(a)}^{12} \cdot {(b)}^{0} + 12 \cdot {(a)}^{11} \cdot {(b)}^{1} + \dots + 1 \cdot {(a)}^{0} \cdot {(b)}^{12}$

Step 4

Simplify the polynomial result.

$a^{12} + 12 a^{11} b + 66 a^{10} b^{2} + 220 a^{9} b^{3} + 495 a^{8} b^{4} + 792 a^{7} b^{5} + 924 a^{6} b^{6} + 792 a^{5} b^{7} + 495 a^{4} b^{8} + 220 a^{3} b^{9} + 66 a^{2} b^{10} + 12 a b^{11} + b^{12}$

Expand Using the Binomial Theorem (a+b)^12

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