# How do you use the Binomial Theorem to expand (5x + 2y)^6 ?

Jan 26, 2016

$15625 {x}^{6} + 37500 {x}^{5} y + 37500 {x}^{4} {y}^{2} + 20000 {x}^{3} {y}^{3} + 6000 {x}^{2} {y}^{4} + 960 x {y}^{5} + 64 {y}^{6}$

#### Explanation:

The Binomial Theorem states that
${\left(a + b\right)}^{n} = {\text{_nC_0a^nb^0 + ""_(n-1)C_1a^(n-1)b^1 + }}_{n - 2} {C}_{2} {a}^{n - 2} {b}^{2} +$..... ""_0C_n a^0b^n
where ""_(n-x)C_x =(n!)/((n-x)!x!)

$\therefore {\left(5 x + 2 y\right)}^{6} =$
(6!)/(6!0!)(5x)^6 +(6!)/(5!1!)(5x)^5(2y)+(6!)/(4!2!) (5x)^4(2y)^2 +(6!)/(3!3!)(5x)^3(2y)^3 + (6!)/(2!4!)(5x)^2(2y)^4 +(6!)/(1!5!)(5x)(2y)^5 +(6!)/(0!6!)(2y)^6

$= 1 \cdot {5}^{6} {x}^{6} + 6 \left({5}^{5}\right) {x}^{5} \cdot 2 y + 15 \cdot {5}^{4} {x}^{4} \cdot {2}^{2} {y}^{2} + 20 \cdot {5}^{3} {x}^{3} \cdot {2}^{3} {y}^{3} + 15 \cdot {5}^{2} {x}^{2} \cdot {2}^{4} {y}^{4} + 6 \cdot 5 x \cdot {2}^{5} {y}^{5} + 1 \cdot {2}^{6} {y}^{6}$

$= 15625 {x}^{6} + 37500 {x}^{5} y + 37500 {x}^{4} {y}^{2} + 20000 {x}^{3} {y}^{3} + 6000 {x}^{2} {y}^{4} + 960 x {y}^{5} + 64 {y}^{6}$