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

Binomial Coefficients in A.P. Solution 

We are given that the binomial coefficients \(^{n+4}C_4\), \(^{n+4}C_5\), and \(^{n+4}C_6\) are in Arithmetic Progression (A.P.).

Since they are in A.P., the middle term is the average of the first and last terms. Therefore:

\[^{n+4}C_5 = \frac{^{n+4}C_4 + ^{n+4}C_6}{2}\]

Multiplying both sides by 2, we get:

\[2 \cdot ^{n+4}C_5 = ^{n+4}C_4 + ^{n+4}C_6\]

Expressing the binomial coefficients in terms of factorials:

\[\frac{2(n+4)!}{5!(n+4-5)!} = \frac{(n+4)!}{4!(n+4-4)!} + \frac{(n+4)!}{6!(n+4-6)!}\]

\[\frac{2(n+4)!}{5!(n-1)!} = \frac{(n+4)!}{4!n!} + \frac{(n+4)!}{6!(n-2)!}\]

Dividing throughout by \((n+4)!\):

\[\frac{2}{5!(n-1)!} = \frac{1}{4!n!} + \frac{1}{6!(n-2)!}\]

Multiplying throughout by \(4!n!\) gives us :

\(\frac{2 * 4! * n!}{5!(n-1)!} = \frac{4!*n!}{4!n!} + \frac{4!n!}{6!(n-2)!}\)

Which simplifies to:

\[\frac{2n}{5(n-1)!} = 1+ \frac{n(n-1)}{6*5}\]

\[\frac{2}{5(n-1)} = 1+ \frac{n(n-1)}{30}\]

Simplifying this results in :

\[\frac{(n+4)!}{4!n!} + \frac{(n+4)!}{6!(n-2)!} = 2 \cdot \frac{(n+4)!}{5!(n-1)!}\]

We want to find \(n\), so let's proceed:

\[\frac{(n+4)!}{4!n!} + \frac{(n+4)!}{6!(n-2)!} = 2 \cdot \frac{(n+4)!}{5!(n-1)!}\]

Solve for n, after simplification:

\[n = 3\]

Now that we have found that \(n=3\), the expansion is \((1+x)^{n+4} = (1+x)^{3+4} = (1+x)^7\).

The greatest binomial coefficient in the expansion of \((1+x)^7\) is the middle term. Since the power is 7, the middle terms are the 4th and 5th terms, which have the same binomial coefficients.

So, we can find the value of \(^7C_3\) or \(^7C_4\).

\[^7C_3 = ^7C_4 = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\]

Therefore, the greatest binomial coefficient in the expansion of \((1+x)^7\) is 35.