How do you use the binomial series to expand  (x + 3)^9 ?

Jun 19, 2017

$\therefore {\left(x + 3\right)}^{9} = {x}^{9} + 27 {x}^{8} + 324 {x}^{7} + 2268 {x}^{6} + 10206 {x}^{5} + 30618 {x}^{4} + 61236 {x}^{3} + 78732 {x}^{2} + 59049 x + 19683$

Explanation:

We know ${\left(a + b\right)}^{n} = n {C}_{0} {a}^{n} \cdot {b}^{0} + n {C}_{1} {a}^{n - 1} \cdot {b}^{1} + n {C}_{2} {a}^{n - 2} \cdot {b}^{2} + \ldots \ldots \ldots . + n {C}_{n} {a}^{n - n} \cdot {b}^{n}$

Here $a = x , b = 3 , n = 9$ We know, nC_r = (n!)/(r!*(n-r)!
$\therefore 9 {C}_{0} = 1 , 9 {C}_{1} = 9 , 9 {C}_{2} = 36 , 9 {C}_{3} = 84 , 9 {C}_{4} = 126 , 9 {C}_{5} = 126 , 9 {C}_{6} = 84 , 9 {C}_{7} = 36 , 9 {C}_{8} = 9 , 9 {C}_{9} = 1$

$\therefore {\left(x + 3\right)}^{9} = {x}^{9} + 9 \cdot {x}^{8} \cdot 3 + 36 \cdot {x}^{7} \cdot {3}^{2} + 84 \cdot {x}^{6} \cdot {3}^{3} + 126 \cdot {x}^{5} \cdot {3}^{4} + 126 \cdot {x}^{4} \cdot {3}^{5} + 84 \cdot {x}^{3} \cdot {3}^{6} + 36 \cdot {x}^{2} \cdot {3}^{7} + 9 x \cdot {3}^{8} + {3}^{9}$ or

$\therefore {\left(x + 3\right)}^{9} = {x}^{9} + 27 {x}^{8} + 324 {x}^{7} + 2268 {x}^{6} + 10206 {x}^{5} + 30618 {x}^{4} + 61236 {x}^{3} + 78732 {x}^{2} + 59049 x + 19683$ [Ans]