Expand Using the Binomial Theorem (x+3/2)^12

${(x + \frac{3}{2})}^{12}$

Step 1

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

$\sum_{k = 0}^{12} \frac{12!}{(12 - k)! k!} \cdot {(x)}^{12 - k} \cdot {(\frac{3}{2})}^{k}$

Step 2

Expand the summation.

$\frac{12!}{(12 - 0)! 0!} {(x)}^{12 - 0} \cdot {(\frac{3}{2})}^{0} + \frac{12!}{(12 - 1)! 1!} {(x)}^{12 - 1} \cdot {(\frac{3}{2})}^{1} + \dots + \frac{12!}{(12 - 12)! 12!} {(x)}^{12 - 12} \cdot {(\frac{3}{2})}^{12}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{12} \cdot {(\frac{3}{2})}^{0} + 12 \cdot {(x)}^{11} \cdot {(\frac{3}{2})}^{1} + \dots + 1 \cdot {(x)}^{0} \cdot {(\frac{3}{2})}^{12}$

Step 4

Simplify the polynomial result.

$x^{12} + 18 x^{11} + \frac{297 x^{10}}{2} + \frac{1485 x^{9}}{2} + \frac{40095 x^{8}}{16} + \frac{24057 x^{7}}{4} + \frac{168399 x^{6}}{16} + \frac{216513 x^{5}}{16} + \frac{3247695 x^{4}}{256} + \frac{1082565 x^{3}}{128} + \frac{1948617 x^{2}}{512} + \frac{531441 x}{512} + \frac{531441}{4096}$

Expand Using the Binomial Theorem (x+3/2)^12

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