# How do you use the binomial (2v+3)^6 using Pascal's triangle?

Apr 9, 2017

$64 {v}^{6} + 576 {v}^{5} + 2160 {v}^{4} + 4320 {v}^{3} + 4860 {v}^{2} + 2916 v + 729$

#### Explanation:

Expand using the corresponding row of coefficients from Pascal's triangle for n = 6

$\Rightarrow 1 \textcolor{w h i t e}{\times} 6 \textcolor{w h i t e}{\times} 15 \textcolor{w h i t e}{\times} 20 \textcolor{w h i t e}{\times} 15 \textcolor{w h i t e}{\times} 6 \textcolor{w h i t e}{\times} 1$

with $\textcolor{b l u e}{\text{decreasing powers" " of "2v" from }} {\left(2 v\right)}^{6} \to {\left(2 v\right)}^{0}$

$\text{and "color(blue)" increasing powers"" of " 3" from } {3}^{0} \to {3}^{6}$

$1. {\left(2 v\right)}^{6.} {3}^{0} + 6. {\left(2 v\right)}^{5.} {3}^{1} + 15. {\left(2 v\right)}^{4.} {3}^{2} + 20. {\left(2 v\right)}^{3.} {3}^{3}$

$+ 15. {\left(2 v\right)}^{2.} {3}^{4} + 6. {\left(2 v\right)}^{1.} {3}^{5} + 1. {\left(2 v\right)}^{0.} {3}^{6}$

$= 64 {v}^{6} + 576 {v}^{5} + 2160 {v}^{4} + 4320 {v}^{3} + 4860 {v}^{2} + 2916 v + 729$