Question

For some \( n \neq 10 \), let the coefficients of the 5^th, 6^th, and 7^th terms in the binomial expansion of \( (1 + x)^{n+4} \) be in A.P. Then the largest coefficient in the expansion of \( (1 + x)^{n+4} \) is:

• A. 70
• B. 35
• C. 20
• D. 10

Hint:

When dealing with binomial expansions, if the coefficients of terms are given to be in arithmetic progression, use the condition \( 2 \binom{n+4}{5} = \binom{n+4}{4} + \binom{n+4}{6} \) to find the value of \( n \). Then calculate the largest coefficient at the middle term.

Answer 1

The Correct Option is B

Step 1: Express the binomial coefficients of the 5th, 6th, and 7th terms. In the expansion of \( (1+x)^{n+4} \), the 5th term is \( \binom{n+4}{4} \), the 6th term is \( \binom{n+4}{5} \), and the 7th term is \( \binom{n+4}{6} \). 
Step 2: Use the property of arithmetic progression. Since the coefficients are in arithmetic progression, we have: \[ 2 \binom{n+4}{5} = \binom{n+4}{4} + \binom{n+4}{6}. \] Step 3: Simplify the equation using the formula for binomial coefficients. Using the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), we can rewrite the equation as: \[ 2 \frac{(n+4)!}{5!(n-1)!} = \frac{(n+4)!}{4!n!} + \frac{(n+4)!}{6!(n-2)!}. \] We can divide by \( (n+4)! \) to get: \[ \frac{2}{5!(n-1)!} = \frac{1}{4!n!} + \frac{1}{6!(n-2)!}. \] Multiplying by \( 6!n! \) gives \[ 2 \cdot \frac{6!n!}{5!(n-1)!} = \frac{6!n!}{4!n!} + \frac{6!n!}{6!(n-2)!}. \] Simplifying: \[ 2 \cdot 6n = 6 \cdot 5 + n(n-1). \] \[ 12n = 30 + n^2 - n \implies n^2 - 13n + 30 = 0. \] Step 4: Solve for \( n \). Factoring the quadratic gives \[ (n-3)(n-10) = 0. \] Since \( n \ne 10 \), we must have \( n = 3 \). 
Step 5: Find the largest coefficient in the expansion of \( (1+x)^{n+4} \). Since \( n = 3 \), the expansion is \( (1+x)^{3+4} = (1+x)^7 \). 
The largest coefficient is the middle term, which is \( \binom{7}{3} = \binom{7}{4} \). 
Step 6: Calculate the largest coefficient. \[ \binom{7}{3} = \frac{7!}{3!4!} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 7 \cdot 5 = 35. \] Final Answer: The largest coefficient in the expansion of \( (1+x)^{n+4} \) is 35.