Use Pascal’s triangle to expand${(2x - 3)^5}$ ?

Hint:The given question requires us to find the expansion of the given binomial expression using Pascal’s triangle and binomial theorem. We can find the value of binomial expansion easily. Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving the given question using the binomial theorem.

Complete step by step answer:
Pascal’s triangle leads to Binomial Expansion. We have to find the binomial expansion of ${(2x - 3)^5}$ . So, using the binomial theorem, the binomial expansion of \[{\left( {x + y} \right)^n}\] is $\sum\nolimits_{r = 0}^n {\left( {^n{C_r}} \right){{\left( x \right)}^{n - r}}{{\left( y \right)}^r}} $.So, the binomial expansion of ${(2x - 3)^5}$ is \[\sum\nolimits_{r = 0}^5 {\left( {^5{C_r}} \right){{\left( {2x} \right)}^{5 - r}}{{\left( { - 3} \right)}^r}} \] .

Now, we have to expand the expression \[\sum\nolimits_{r = 0}^5 {\left( {^5{C_r}} \right){{\left( {2x} \right)}^{5 - r}}{{\left( { - 3} \right)}^r}} \] and we are done with the binomial expansion of ${(2x - 3)^5}$ .
\[\sum\nolimits_{r = 0}^5 {\left( {^5{C_r}} \right){{\left( {2x} \right)}^{5 - r}}{{\left( { - 3} \right)}^r}} = \left\{\left( {^5{C_0}} \right){\left( {2x} \right)^5}{\left( { - 3} \right)^0} + \left( {^5{C_1}} \right){\left( {2x} \right)^4}{\left( { - 3} \right)^1} + \left( {^5{C_2}} \right){\left( {2x} \right)^3}{\left( { - 3} \right)^2} \\
+ \left( {^5{C_3}} \right){\left( {2x} \right)^2}{\left( { - 3} \right)^3} + \left( {^5{C_4}} \right){\left( {2x} \right)^1}{\left( { - 3} \right)^4} + \left( {^5{C_5}} \right){\left( {2x} \right)^0}{\left( { - 3} \right)^5}\right\}\]

Substituting the values of factorials, we get,
\[\sum\nolimits_{r = 0}^5 {\left( {^5{C_r}} \right){{\left( {2x} \right)}^{5 - r}}{{\left( { - 3} \right)}^r}} = \left\{{\left( {2x} \right)^5}{\left( { - 3} \right)^0} + \left( 5 \right){\left( {2x} \right)^4}{\left( { - 3} \right)^1} + \left( {10} \right){\left( {2x} \right)^3}{\left( { - 3} \right)^2} \\
+ \left( {10} \right){\left( {2x} \right)^2}{\left( { - 3} \right)^3} + \left( 5 \right){\left( {2x} \right)^1}{\left( { - 3} \right)^4} + {\left( {2x} \right)^0}{\left( { - 3} \right)^5}\right\}\]

Calculating the values of all the terms raised to the powers, we get,
\[\sum\nolimits_{r = 0}^5 {\left( {^5{C_r}} \right){{\left( {2x} \right)}^{5 - r}}{{\left( { - 3} \right)}^r}} = \left\{ 32{x^5} + \left( 5 \right)\left( {16{x^4}} \right)\left( { - 3} \right) + \left( {10} \right)\left( {8{x^3}} \right)\left( 9 \right) \\
+ \left( {10} \right)\left( {4{x^2}} \right)\left( { - 27} \right) + \left( 5 \right)\left( {2x} \right)\left( {81} \right) + \left( 1 \right)\left( { - 243} \right) \right\}\]

Calculating the products and doing calculations, we get,
\[\sum\nolimits_{r = 0}^5 {\left( {^5{C_r}} \right){{\left( {2x} \right)}^{5 - r}}{{\left( { - 3} \right)}^r}} = 32{x^5} - 240{x^4} + 720{x^3} - 1080{x^2} + 810x - 243\]
Hence, the value of \[\sum\nolimits_{r = 0}^5 {\left( {^5{C_r}} \right){{\left( {2x} \right)}^{5 - r}}{{\left( { - 3} \right)}^r}} \] is \[\left( {32{x^5} - 240{x^4} + 720{x^3} - 1080{x^2} + 810x - 243} \right)\] .

So, Binomial expansion of ${(2x - 3)^5}$ is \[\left( {32{x^5} - 240{x^4} + 720{x^3} - 1080{x^2} + 810x - 243} \right)\].

Note:The given problem can be solved by various methods. The easiest way is to apply the concepts of Pascal’s Triangle and Binomial theorem as it is very effective in finding the binomial expansion. The coefficients of the terms of binomial expansion are the building blocks of Pascal's triangle.