Expand Using the Binomial Theorem (x+1)^9

${(x + 1)}^{9}$

Step 1

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

$\sum_{k = 0}^{9} \frac{9!}{(9 - k)! k!} \cdot {(x)}^{9 - k} \cdot {(1)}^{k}$

Step 2

Expand the summation.

$\frac{9!}{(9 - 0)! 0!} {(x)}^{9 - 0} \cdot {(1)}^{0} + \frac{9!}{(9 - 1)! 1!} {(x)}^{9 - 1} \cdot {(1)}^{1} + \dots + \frac{9!}{(9 - 9)! 9!} {(x)}^{9 - 9} \cdot {(1)}^{9}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{9} \cdot {(1)}^{0} + 9 \cdot {(x)}^{8} \cdot {(1)}^{1} + \dots + 1 \cdot {(x)}^{0} \cdot {(1)}^{9}$

Step 4

Simplify the polynomial result.

$x^{9} + 9 x^{8} + 36 x^{7} + 84 x^{6} + 126 x^{5} + 126 x^{4} + 84 x^{3} + 36 x^{2} + 9 x + 1$

Expand Using the Binomial Theorem (x+1)^9

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