Expandera $$$\left(3 x^{2} + y\right)^{5}$$$

Utveckla $$$\left(3 x^{2} + y\right)^{5}$$$.

Utvecklingen ges av följande formel: $$$\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$$$, där $$${\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!}$$$ och $$$n! = 1 \cdot 2 \cdot \ldots \cdot n$$$.

Vi har att $$$a = 3 x^{2}$$$, $$$b = y$$$ och $$$n = 5$$$.

Således, $$$\left(3 x^{2} + y\right)^{5} = \sum_{k=0}^{5} {\binom{5}{k}} \left(3 x^{2}\right)^{5 - k} y^{k}$$$.

Beräkna nu produkten för varje värde på $$$k$$$ från $$$0$$$ till $$$5$$$.

$$$k = 0$$$: $$${\binom{5}{0}} \left(3 x^{2}\right)^{5 - 0} y^{0} = \frac{5!}{\left(5 - 0\right)! 0!} \left(3 x^{2}\right)^{5 - 0} y^{0} = 243 x^{10}$$$

$$$k = 1$$$: $$${\binom{5}{1}} \left(3 x^{2}\right)^{5 - 1} y^{1} = \frac{5!}{\left(5 - 1\right)! 1!} \left(3 x^{2}\right)^{5 - 1} y^{1} = 405 x^{8} y$$$

$$$k = 2$$$: $$${\binom{5}{2}} \left(3 x^{2}\right)^{5 - 2} y^{2} = \frac{5!}{\left(5 - 2\right)! 2!} \left(3 x^{2}\right)^{5 - 2} y^{2} = 270 x^{6} y^{2}$$$

$$$k = 3$$$: $$${\binom{5}{3}} \left(3 x^{2}\right)^{5 - 3} y^{3} = \frac{5!}{\left(5 - 3\right)! 3!} \left(3 x^{2}\right)^{5 - 3} y^{3} = 90 x^{4} y^{3}$$$

$$$k = 4$$$: $$${\binom{5}{4}} \left(3 x^{2}\right)^{5 - 4} y^{4} = \frac{5!}{\left(5 - 4\right)! 4!} \left(3 x^{2}\right)^{5 - 4} y^{4} = 15 x^{2} y^{4}$$$

$$$k = 5$$$: $$${\binom{5}{5}} \left(3 x^{2}\right)^{5 - 5} y^{5} = \frac{5!}{\left(5 - 5\right)! 5!} \left(3 x^{2}\right)^{5 - 5} y^{5} = y^{5}$$$

Alltså, $$$\left(3 x^{2} + y\right)^{5} = 243 x^{10} + 405 x^{8} y + 270 x^{6} y^{2} + 90 x^{4} y^{3} + 15 x^{2} y^{4} + y^{5}.$$$

$$$\left(3 x^{2} + y\right)^{5} = 243 x^{10} + 405 x^{8} y + 270 x^{6} y^{2} + 90 x^{4} y^{3} + 15 x^{2} y^{4} + y^{5}$$$A