Question

How do you use the binomial theorem to expand \[{\left( {1 + 3x} \right)^{\dfrac{3}{2}}}\] ?

Answer 1

Hint: The given question requires us to find the binomial expansion of the given binomial expression \[{\left( {1 + 3x} \right)^{\dfrac{3}{2}}}\] . We can find the required binomial expansion by using the Binomial theorem. 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:
We have to find the binomial expansion of \[{\left( {1 + 3x} \right)^{\dfrac{3}{2}}}\] . So, using the binomial theorem, the binomial expansion of \[{\left( {1 + x} \right)^n}\] is $1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + ....$
So, the binomial expansion of \[{\left( {1 + 3x} \right)^{\dfrac{3}{2}}}\] can be calculated by the above expression by substituting in the correct value of n and x.
Now, we compare the expression \[{\left( {1 + 3x} \right)^{\dfrac{3}{2}}}\] with \[{\left( {1 + x} \right)^n}\] . So, we get the value of n as $\left( {\dfrac{3}{2}} \right)$ and 3x in place of x.
So, we have,
 \[ = 1 + \left( {\dfrac{3}{2}} \right)\left( {3x} \right) + \dfrac{{\left( {\dfrac{3}{2}} \right)\left( {\left( {\dfrac{3}{2}} \right) - 1} \right)}}{{2!}}{\left( {3x} \right)^2} + \dfrac{{\left( {\dfrac{3}{2}} \right)\left( {\left( {\dfrac{3}{2}} \right) - 1} \right)\left( {\left( {\dfrac{3}{2}} \right) - 2} \right)}}{{3!}}{\left( {3x} \right)^3} + \dfrac{{\left( {\dfrac{3}{2}} \right)\left( {\left( {\dfrac{3}{2}} \right) - 1} \right)\left( {\left( {\dfrac{3}{2}} \right) - 2} \right)\left( {\left( {\dfrac{3}{2}} \right) - 3} \right)}}{{4!}}{\left( {3x} \right)^4} + ....\]
So, simplifying the expression further, we get,
 \[ = 1 + \left( {\dfrac{3}{2}} \right)\left( {3x} \right) + \dfrac{{\left( {\dfrac{3}{2}} \right)\left( {\dfrac{1}{2}} \right)}}{{2!}}{\left( {3x} \right)^2} + \dfrac{{\left( {\dfrac{3}{2}} \right)\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)}}{{3!}}{\left( {3x} \right)^3} + \dfrac{{\left( {\dfrac{3}{2}} \right)\left( {\dfrac{1}{2}} \right)\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)}}{{4!}}{\left( {3x} \right)^4} + ....\]
So, we can evaluate first few terms of the binomial expansion,
 \[ = 1 + \dfrac{9}{2}x + \dfrac{{\left( {\dfrac{3}{4}} \right)}}{2}{\left( {3x} \right)^2} + \dfrac{{\left( { - \dfrac{3}{8}} \right)}}{6}{\left( {3x} \right)^3} + \dfrac{{\left( {\dfrac{9}{{16}}} \right)}}{{24}}{\left( {3x} \right)^4} + ....\]
Simplifying the expression and opening the brackets, we get,
 \[ = 1 + \dfrac{9}{2}x + \dfrac{3}{8} \times 9{x^2} + \dfrac{{ - 1}}{{16}} \times 27{x^3} + \dfrac{3}{{128}} \times 81{x^4} + ....\]
Simplifying the expression further,
 \[ = 1 + \dfrac{9}{2}x + \dfrac{{27}}{8}{x^2} - \dfrac{{27}}{{16}}{x^3} + \dfrac{{243}}{{128}}{x^4} + ....\]
Hence, the expression \[{\left( {1 + 3x} \right)^{\dfrac{3}{2}}}\] can be simplified as \[ = 1 + \dfrac{9}{2}x + \dfrac{{27}}{8}{x^2} - \dfrac{{27}}{{16}}{x^3} + \dfrac{{243}}{{128}}{x^4} + ....\] using binomial theorem.
So, the correct answer is “ \[ = 1 + \dfrac{9}{2}x + \dfrac{{27}}{8}{x^2} - \dfrac{{27}}{{16}}{x^3} + \dfrac{{243}}{{128}}{x^4} + ....\] ”.

Note: The given problem can be solved by various methods. The easiest way is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. One should be careful while carrying out the calculations and simplification.