Use Pascal’s Triangle to expand${(2x-3)}^4$

We are given the expression :

${(2x-3)}^4$and we need to solve this using the Pascal's triangle

First we need to expand the given expression by using the formula :

${(a+b)}^n=\_\_a^n+\_\_a^{n-1}{b}^1+\_\_a^{n-2}{b}^2+...+\_\_a^1{b}^{n-1}+\_\_b^n$

After this we find the co-efficient by using the Pascal's triangle , we count the number of expression obtained and then use that number as the row number of the triangle :

We need to expand the given expression :

${(2x-3)}^4=\_\_2x^4+\_\_2x^{4-1}{(-3)}^1+\_\_2x^{4-2}{(-3)}^2+\_\_2x^{4-3}{(-3)}^3+\_\_2x^{4-4}{(-3)}^4$

role="math" localid="1645460031687" ${(2x-3)}^4=\_\_16x^4+\_\_-24x^3+\_\_36x^2+\_\_-54x+81$

We see that the number of expression obtained after expanding is six so we will take the co-efficient from the sixth row of the Pascal's triangle, so we will get :

$16x^4-4.24x^3+6.36x^2-4.54x+81=16x^4-96x^3+216x^2-216x+81$