Expand Using the Binomial Theorem (2/a+(a^2)/3)^11

${(\frac{2}{a} + \frac{a^{2}}{3})}^{11}$

Step 1

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

$\sum_{k = 0}^{11} \frac{11!}{(11 - k)! k!} \cdot {(\frac{2}{a})}^{11 - k} \cdot {(\frac{a^{2}}{3})}^{k}$

Step 2

Expand the summation.

$\frac{11!}{(11 - 0)! 0!} \cdot {(\frac{2}{a})}^{11 - 0} \cdot {(\frac{a^{2}}{3})}^{0} + \frac{11!}{(11 - 1)! 1!} \cdot {(\frac{2}{a})}^{11 - 1} \cdot {(\frac{a^{2}}{3})}^{1} + \dots + \frac{11!}{(11 - 11)! 11!} \cdot {(\frac{2}{a})}^{11 - 11} \cdot {(\frac{a^{2}}{3})}^{11}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(\frac{2}{a})}^{11} \cdot {(\frac{a^{2}}{3})}^{0} + 11 \cdot {(\frac{2}{a})}^{10} \cdot {(\frac{a^{2}}{3})}^{1} + \dots + 1 \cdot {(\frac{2}{a})}^{0} \cdot {(\frac{a^{2}}{3})}^{11}$

Step 4

Simplify the polynomial result.

$\frac{2048}{a^{11}} + \frac{11264}{3 a^{8}} + \frac{28160}{9 a^{5}} + \frac{14080}{9 a^{2}} + \frac{14080 a}{27} + \frac{9856 a^{4}}{81} + \frac{4928 a^{7}}{243} + \frac{1760 a^{10}}{729} + \frac{440 a^{13}}{2187} + \frac{220 a^{16}}{19683} + \frac{22 a^{19}}{59049} + \frac{a^{22}}{177147}$

Expand Using the Binomial Theorem (2/a+(a^2)/3)^11

Related Questions

Support LCMGCF.com by disabling your adblocker.