How do you use the binomial theorem to expand and simplify the expression (x+y)^5?

Jan 29, 2018

$\textcolor{g r e e n}{{\left(x + y\right)}^{5} = {x}^{5} + 5 {x}^{4} y + 10 {x}^{3} {y}^{2} + 10 {x}^{2} {y}^{3} + 5 x {y}^{4} + {y}^{5}}$

Explanation:

${\left(x + y\right)}^{5} = 5 C 0 {x}^{5} {y}^{0} + 5 C 1 {x}^{4} {y}^{1} + 5 C 2 {x}^{3} {y}^{2} + 5 C 3 {x}^{2} {y}^{3} + 5 C 4 x {y}^{4} + 5 C 5 {x}^{0} {y}^{5}$

$= {x}^{5} + 5 {x}^{4} y + \left(\frac{5 \cdot 4}{1 \cdot 2}\right) \cdot {x}^{3} {y}^{2} + \left(\frac{5 \cdot 4 \cdot 3}{1 \cdot 2 \cdot 3}\right) {x}^{2} {y}^{3} + \left(\frac{5 \cdot 4 \cdot 3 \cdot 2}{1 \cdot 2 \cdot 3 \cdot 4}\right) x {y}^{4} + {y}^{5}$

$\textcolor{g r e e n}{= {x}^{5} + 5 {x}^{4} y + 10 {x}^{3} {y}^{2} + 10 {x}^{2} {y}^{3} + 5 x {y}^{4} + {y}^{5}}$

Jan 29, 2018

$\textcolor{b l u e}{{\left(x + y\right)}^{5} = {x}^{5} + 5 {x}^{4} y + 10 {x}^{3} {y}^{2} + 10 {x}^{2} {y}^{3} + 5 x {y}^{4} + {y}^{5}}$

Explanation:

Alternate Method using Pascal's Triangle,

1 5 10 10 5 1

$\textcolor{b l u e}{{\left(x + y\right)}^{5} = {x}^{5} + 5 {x}^{4} y + 10 {x}^{3} {y}^{2} + 10 {x}^{2} {y}^{3} + 5 x {y}^{4} + {y}^{5}}$