Expand Using the Binomial Theorem (2x-1)^11

${(2 x - 1)}^{11}$

Step 1

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

$\sum_{k = 0}^{11} \frac{11!}{(11 - k)! k!} \cdot {(2 x)}^{11 - k} \cdot {(- 1)}^{k}$

Step 2

Expand the summation.

$\frac{11!}{(11 - 0)! 0!} \cdot {(2 x)}^{11 - 0} \cdot {(- 1)}^{0} + \frac{11!}{(11 - 1)! 1!} \cdot {(2 x)}^{11 - 1} \cdot {(- 1)}^{1} + \dots + \frac{11!}{(11 - 11)! 11!} \cdot {(2 x)}^{11 - 11} \cdot {(- 1)}^{11}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 x)}^{11} \cdot {(- 1)}^{0} + 11 \cdot {(2 x)}^{10} \cdot {(- 1)}^{1} + \dots + 1 \cdot {(2 x)}^{0} \cdot {(- 1)}^{11}$

Step 4

Simplify the polynomial result.

$2048 x^{11} - 11264 x^{10} + 28160 x^{9} - 42240 x^{8} + 42240 x^{7} - 29568 x^{6} + 14784 x^{5} - 5280 x^{4} + 1320 x^{3} - 220 x^{2} + 22 x - 1$

Expand Using the Binomial Theorem (2x-1)^11

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