Expand Using the Binomial Theorem (3/y+y^2)^5

${(\frac{3}{y} + y^{2})}^{5}$

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

$\sum_{k = 0}^{5} \frac{5!}{(5 - k)! k!} \cdot {(\frac{3}{y})}^{5 - k} \cdot {(y^{2})}^{k}$

Expand the summation.

$\frac{5!}{(5 - 0)! 0!} \cdot {(\frac{3}{y})}^{5 - 0} \cdot {(y^{2})}^{0} + \frac{5!}{(5 - 1)! 1!} \cdot {(\frac{3}{y})}^{5 - 1} \cdot {(y^{2})}^{1} + \dots + \frac{5!}{(5 - 5)! 5!} \cdot {(\frac{3}{y})}^{5 - 5} \cdot {(y^{2})}^{5}$

Simplify the exponents for each term of the expansion.

$1 \cdot {(\frac{3}{y})}^{5} \cdot {(y^{2})}^{0} + 5 \cdot {(\frac{3}{y})}^{4} \cdot {(y^{2})}^{1} + \dots + 1 \cdot {(\frac{3}{y})}^{0} \cdot {(y^{2})}^{5}$

Simplify the polynomial result.

$\frac{243}{y^{5}} + \frac{405}{y^{2}} + 270 y + 90 y^{4} + 15 y^{7} + y^{10}$

Expand Using the Binomial Theorem (3/y+y^2)^5

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