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Question: Find the binomial expansion of \[{\left( {2x + 1} \right)^3}\] ....

Find the binomial expansion of (2x+1)3{\left( {2x + 1} \right)^3} .

Explanation

Solution

The given question requires us to find the binomial expansion of the given binomial expression. 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 (2x+1)3{\left( {2x + 1} \right)^3} . So, using the binomial theorem, the binomial expansion of (x+y)n{\left( {x + y} \right)^n} is ∑r=0n(nCr)(x)n−r(y)r\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+1)3{\left( {2x + 1} \right)^3} is ∑r=03(3Cr)(2x)3−r(1)r\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( 1 \right)}^r}} .

Now, we have to expand the expression ∑r=03(3Cr)(2x)3−r(1)r\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( 1 \right)}^r}} and we are done with the binomial expansion of (2x+1)3{\left( {2x + 1} \right)^3} .

∑r=03(3Cr)(2x)3−r(1)r=(3C0)(2x)3(1)0+(3C1)(2x)2(1)1+(3C2)(2x)1(1)2+(3C3)(2x)0(1)3\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( 1 \right)}^r}} = \left( {^3{C_0}} \right){\left( {2x} \right)^3}{\left( 1 \right)^0} + \left( {^3{C_1}} \right){\left( {2x} \right)^2}{\left( 1 \right)^1} + \left( {^3{C_2}} \right){\left( {2x} \right)^1}{\left( 1 \right)^2} + \left( {^3{C_3}} \right){\left( {2x} \right)^0}{\left( 1 \right)^3}

Since any power of one is equal to one itself. So, equating all the brackets with powers of 11 as 11.

Hence, ∑r=03(3Cr)(2x)3−r(1)r=(3C0)(2x)3+(3C1)(2x)2+(3C2)(2x)1+(3C3)(2x)0\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( 1 \right)}^r}} = \left( {^3{C_0}} \right){\left( {2x} \right)^3} + \left( {^3{C_1}} \right){\left( {2x} \right)^2} + \left( {^3{C_2}} \right){\left( {2x} \right)^1} + \left( {^3{C_3}} \right){\left( {2x} \right)^0} .

Substituting values of combination formulae and simplifying calculations, we get

=∑r=03(3Cr)(2x)3−r(1)r=(1)(8x3)+(3)(4x2)+(3)(2x)+(1)(1) = \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( 1 \right)}^r}} = \left( 1 \right)\left( {8{x^3}} \right) + \left( 3 \right)\left( {4{x^2}} \right) + \left( 3 \right)\left( {2x} \right) + \left( 1 \right)\left( 1 \right)

Opening brackets and doing calculations, we get

=∑r=03(3Cr)(2x)3−r(1)r=(8x3)+(12x2)+(6x)+(1) = \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( 1 \right)}^r}} = \left( {8{x^3}} \right) + \left( {12{x^2}} \right) + \left( {6x} \right) + \left( 1 \right)

=∑r=03(3Cr)(2x)3−r(1)r=(8x3+12x2+6x+1) = \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( 1 \right)}^r}} = \left( {8{x^3} + 12{x^2} + 6x + 1} \right)

Hence, the value of ∑r=03(3Cr)(2x)3−r(1)r\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( 1 \right)}^r}} is (8x3+12x2+6x+1)\left( {8{x^3} + 12{x^2} + 6x + 1} \right) .

So, Binomial expansion of (2x+1)3{\left( {2x + 1} \right)^3} is (8x3+12x2+6x+1)\left( {8{x^3} + 12{x^2} + 6x + 1} \right) .

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. But the problem can also be solved by actually calculating the cube of the binomial expansion given, though it is a tedious task to calculate the cube of a binomial polynomial.