Given that in the expansion of 

(4x2−3x)n

, the 

5th

 term, in descending powers of 

x

, is a constant, find the value of 

n

.

To find the value of \( n \) in the expansion of \((4x^2 - 3x)^n\) such that the 5th term is a constant, we need to determine the term which corresponds to the 5th position in the binomial expansion.

The general term \( T_k \) in the binomial expansion of \((a + b)^n\) is given by:
\[
T_k = \binom{n}{k} a^{n-k} b^k
\]
where \( a = 4x^2 \) and \( b = -3x \).

To find the 5th term (which is actually \( T_4 \) in binomial terms, because indexing starts from \( k=0 \)), we set up the term:
\[
T_4 = \binom{n}{4} (4x^2)^{n-4} (-3x)^4
\]

Simplifying \( T_4 \):
\[
T_4 = \binom{n}{4} (4x^2)^{n-4} (-3x)^4
= \binom{n}{4} (4^{n-4} x^{2(n-4)}) ((-3)^4 x^4)
= \binom{n}{4} 4^{n-4} (-3)^4 x^{2(n-4) + 4}
= \binom{n}{4} 4^{n-4} 81 x^{2n-8 + 4}
= \binom{n}{4} 4^{n-4} 81 x^{2n - 4}
\]

We want this term to be a constant. For it to be a constant, the power of \( x \) must be zero:
\[
2n - 4 = 0
\]

Solving for \( n \):
\[
2n - 4 = 0 \\
2n = 4 \\
n = 2
\]

Therefore, the value of \( n \) is:
\[
\boxed{2}
\]

Would you like more details or have any questions about this solution?

Here are 5 related questions you might want to ask:
1. How do you determine the general term in a binomial expansion?
2. What are the steps to find a specific term in the binomial expansion?
3. How do you handle binomial expansions with negative exponents?
4. What is the binomial theorem and how is it applied?
5. Can you show an example of finding a specific term in a different binomial expansion?

**Tip:** When dealing with binomial expansions, always remember to consider the indexing properly and simplify each term step-by-step to avoid mistakes.

Discover how to find the value of n in the binomial expansion (4x^2 - 3x)^n such that the 5th term is a constant. Learn step-by-step with detailed formulas and calculations. Suitable for high school students studying algebra and binomial expansions.

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