Find the middle term in the expansion of $(2 x-y)^4$ and simplify your answer.
Find the term that contains $x^5$ in the expansion of $(2 x+3 y)^8$.
Answer (2)
Find the middle term of ( 2 x − y ) 4 using the binomial theorem with k = 2 : T 3 = ( 2 4 ) ( 2 x ) 2 ( − y ) 2 = 6 ( 4 x 2 ) ( y 2 ) = 24 x 2 y 2 .
Find the term containing x 5 in ( 2 x + 3 y ) 8 . Determine k such that 8 − k = 5 , which gives k = 3 .
Substitute k = 3 into the general term: T 4 = ( 3 8 ) ( 2 x ) 5 ( 3 y ) 3 = 56 × ( 32 x 5 ) × ( 27 y 3 ) .
Simplify to find the term: T 4 = 48384 x 5 y 3 . The middle term is 24 x 2 y 2 and the term containing x 5 is 48384 x 5 y 3 . 24 x 2 y 2 , 48384 x 5 y 3
Explanation
Understanding the Problem We are given two binomial expressions and asked to find specific terms in their expansions. We will use the binomial theorem to solve this problem. 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 ! is the binomial coefficient.
Finding the Middle Term For part 4.1, we need to find the middle term in the expansion of ( 2 x − y ) 4 . Since the power is 4, there are 5 terms in the expansion (corresponding to k = 0 , 1 , 2 , 3 , 4 ). The middle term is the third term, which corresponds to k = 2 .
Applying the Binomial Theorem Using the binomial theorem, the middle term is given by: T 3 = ( 2 4 ) ( 2 x ) 4 − 2 ( − y ) 2
Simplifying the Expression Now, we simplify the expression: ( 2 4 ) = 2 ! 2 ! 4 ! = ( 2 × 1 ) ( 2 × 1 ) 4 × 3 × 2 × 1 = 6 ( 2 x ) 2 = 4 x 2 ( − y ) 2 = y 2 So, the middle term is: T 3 = 6 ( 4 x 2 ) ( y 2 ) = 24 x 2 y 2
Finding the Term with x^5 For part 4.2, we need to find the term that contains x 5 in the expansion of ( 2 x + 3 y ) 8 . The general term in the binomial expansion is given by: T k + 1 = ( k 8 ) ( 2 x ) 8 − k ( 3 y ) k
Determining the Value of k We want the term with x 5 , so we need to find the value of k such that the power of x is 5. That is, 8 − k = 5 , which gives k = 3 .
Substituting k into the General Term Now, we substitute k = 3 into the general term: T 3 + 1 = T 4 = ( 3 8 ) ( 2 x ) 8 − 3 ( 3 y ) 3
Simplifying the Expression We simplify the expression: ( 3 8 ) = 3 ! 5 ! 8 ! = ( 3 × 2 × 1 ) × 5 ! 8 × 7 × 6 × 5 ! = 3 × 2 × 1 8 × 7 × 6 = 56 ( 2 x ) 5 = 32 x 5 ( 3 y ) 3 = 27 y 3 So, the term containing x 5 is: T 4 = 56 × ( 32 x 5 ) × ( 27 y 3 ) = 56 × 32 × 27 x 5 y 3 = 48384 x 5 y 3
Final Answer Therefore, the middle term in the expansion of ( 2 x − y ) 4 is 24 x 2 y 2 , and the term containing x 5 in the expansion of ( 2 x + 3 y ) 8 is 48384 x 5 y 3 .
Examples
Binomial expansions are used in various fields such as probability, statistics, and physics. For example, in probability, when calculating the likelihood of a certain number of successes in a series of independent trials, the binomial theorem is applied. In physics, it can be used to approximate complex equations, making them easier to solve. Understanding binomial expansions allows us to model and predict outcomes in scenarios involving repeated independent events or approximate solutions in complex systems.
The middle term in the expansion of ( 2 x − y ) 4 is 24 x 2 y 2 , while the term containing x 5 in the expansion of ( 2 x + 3 y ) 8 is 48384 x 5 y 3 .
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