Expand Using the Binomial Theorem (x^3+1/(2x))^15

${(x^{3} + \frac{1}{2 x})}^{15}$

Step 1

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

$\sum_{k = 0}^{15} \frac{15!}{(15 - k)! k!} \cdot {(x^{3})}^{15 - k} \cdot {(\frac{1}{2 x})}^{k}$

Step 2

Expand the summation.

$\frac{15!}{(15 - 0)! 0!} \cdot {(x^{3})}^{15 - 0} \cdot {(\frac{1}{2 x})}^{0} + \frac{15!}{(15 - 1)! 1!} \cdot {(x^{3})}^{15 - 1} \cdot {(\frac{1}{2 x})}^{1} + \dots + \frac{15!}{(15 - 15)! 15!} \cdot {(x^{3})}^{15 - 15} \cdot {(\frac{1}{2 x})}^{15}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x^{3})}^{15} \cdot {(\frac{1}{2 x})}^{0} + 15 \cdot {(x^{3})}^{14} \cdot {(\frac{1}{2 x})}^{1} + \dots + 1 \cdot {(x^{3})}^{0} \cdot {(\frac{1}{2 x})}^{15}$

Step 4

Simplify the polynomial result.

$x^{45} + \frac{15 x^{41}}{2} + \frac{105 x^{37}}{4} + \frac{455 x^{33}}{8} + \frac{1365 x^{29}}{16} + \frac{3003 x^{25}}{32} + \frac{5005 x^{21}}{64} + \frac{6435 x^{17}}{128} + \frac{6435 x^{13}}{256} + \frac{5005 x^{9}}{512} + \frac{3003 x^{5}}{1024} + \frac{1365 x}{2048} + \frac{455}{4096 x^{3}} + \frac{105}{8192 x^{7}} + \frac{15}{16384 x^{11}} + \frac{1}{32768 x^{15}}$

Expand Using the Binomial Theorem (x^3+1/(2x))^15

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