Question

If the coefficients of \( x^4 \), \( x^5 \), and \( x^6 \) in the expansion of \( (1 + x)^n \) are in arithmetic progression, then the maximum value of \( n \) is:

• A. 14
• B. 21
• C. 28
• D. 7

Answer 1

The Correct Option is A

In the binomial expansion of \( (1+x)^n \), the general term is given by \( T_k = \binom{n}{k}x^k \). Therefore, the coefficients of \( x^4 \), \( x^5 \), and \( x^6 \) are:

• Coefficient of \( x^4 \): \( \binom{n}{4} \),
• Coefficient of \( x^5 \): \( \binom{n}{5} \),
• Coefficient of \( x^6 \): \( \binom{n}{6} \).

Since these coefficients are in an arithmetic progression, we can set up the condition:

\[ 2\binom{n}{5} = \binom{n}{4} + \binom{n}{6}. \]

Using the formula for binomial coefficients, we have:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!}. \]

After simplifying, we substitute and solve for \( n \) to find that the maximum value of \( n \) that satisfies this condition is \( n = 14 \).

Therefore, the maximum value of \( n \) is \( 14 \).

Answer 2

Step 1: Understand the problem.
We are given that the coefficients of \( x^4 \), \( x^5 \), and \( x^6 \) in the expansion of \( (1 + x)^n \) form an arithmetic progression. We need to find the maximum value of \( n \).

Step 2: General term in the binomial expansion.
The binomial expansion of \( (1 + x)^n \) is given by:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \] The coefficient of \( x^k \) is \( \binom{n}{k} \). Therefore, the coefficients of \( x^4 \), \( x^5 \), and \( x^6 \) are:
- Coefficient of \( x^4 \) is \( \binom{n}{4} \),
- Coefficient of \( x^5 \) is \( \binom{n}{5} \),
- Coefficient of \( x^6 \) is \( \binom{n}{6} \).
We are told that these coefficients are in arithmetic progression, so:
\[ 2 \cdot \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \] This gives us the equation:
\[ 2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \] Now, let's express the binomial coefficients in terms of \( n \).

Step 3: Expand the binomial coefficients.
Using the properties of binomial coefficients:
- \( \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24} \),
- \( \binom{n}{5} = \frac{n(n-1)(n-2)(n-3)(n-4)}{120} \),
- \( \binom{n}{6} = \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{720} \).
Substitute these into the equation \( 2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \). After simplifying, solving this equation gives the value of \( n = 14 \).

Step 4: Conclusion.
Thus, the maximum value of \( n \) is \( \boxed{14} \).