Apply the binomial theorem to expand the expression.
Calculate each term in the expansion using the formula ( k n ) a n − k b k .
Simplify each term to obtain the coefficients and powers of x .
Combine the terms to get the final expansion: 64 x 6 + 128 x 5 + 3 320 x 4 + 27 1280 x 3 + 27 320 x 2 + 81 128 x + 729 64
Explanation
Understanding the problem We are asked to expand the expression ( 2 x + 3 2 ) 6 using the Binomial Theorem.
Stating the Binomial Theorem The binomial theorem states that ( a + b ) n = ∑ k = 0 n ( k n ) a n − k b k , where ( k n ) = k ! ( n − k )! n !
Applying the theorem to our problem In our case, a = 2 x , b = 3 2 , and n = 6 . So, we need to expand ( 2 x + 3 2 ) 6 = ∑ k = 0 6 ( k 6 ) ( 2 x ) 6 − k ( 3 2 ) k .
Calculating each term Let's calculate the binomial coefficients and simplify each term:
For k = 0 : ( 0 6 ) ( 2 x ) 6 ( 3 2 ) 0 = 1 ⋅ ( 64 x 6 ) ⋅ 1 = 64 x 6
For k = 1 : ( 1 6 ) ( 2 x ) 5 ( 3 2 ) 1 = 6 ⋅ ( 32 x 5 ) ⋅ 3 2 = 128 x 5
For k = 2 : ( 2 6 ) ( 2 x ) 4 ( 3 2 ) 2 = 15 ⋅ ( 16 x 4 ) ⋅ 9 4 = 9 960 x 4 = 3 320 x 4 ≈ 106.67 x 4
For k = 3 : ( 3 6 ) ( 2 x ) 3 ( 3 2 ) 3 = 20 ⋅ ( 8 x 3 ) ⋅ 27 8 = 27 1280 x 3 ≈ 47.41 x 3
For k = 4 : ( 4 6 ) ( 2 x ) 2 ( 3 2 ) 4 = 15 ⋅ ( 4 x 2 ) ⋅ 81 16 = 81 960 x 2 = 27 320 x 2 ≈ 11.85 x 2
For k = 5 : ( 5 6 ) ( 2 x ) 1 ( 3 2 ) 5 = 6 ⋅ ( 2 x ) ⋅ 243 32 = 243 384 x = 81 128 x ≈ 1.58 x
For k = 6 : ${6 \choose 6} (2x)^0 \left(\frac{2}{3}\right)^6 = 1 \cdot 1 \cdot \frac{64}{729} = \frac{64}{729} \approx 0.09
Final Expansion Therefore, the expansion is:
( 2 x + 3 2 ) 6 = 64 x 6 + 128 x 5 + 3 320 x 4 + 27 1280 x 3 + 27 320 x 2 + 81 128 x + 729 64
Examples
The binomial theorem is used in probability theory to calculate the likelihood of a certain number of successes in a series of independent trials. For example, if you flip a coin 6 times, the binomial theorem can help you calculate the probability of getting exactly 3 heads. It also has applications in finance, such as in option pricing models, and in physics, for approximating complex equations.