Answer :
To solve this problem, let's consider the binomial expansion of [tex]\((1 + x)^n\)[/tex], where the [tex]\( k \)[/tex]-th term is given by the binomial coefficient multiplied by [tex]\( x \)[/tex]. Specifically, the [tex]\( (k+1) \)[/tex]-th term, which corresponds to the [tex]\( k \)[/tex]-th power of [tex]\( x \)[/tex], can be written as:
[tex]\[ T_{k+1} = \binom{n}{k} \cdot x^k \][/tex]
Here, we focus on the coefficients of the 5th, 6th, and 7th terms, which correspond to [tex]\( k = 4 \)[/tex], [tex]\( k = 5 \)[/tex], and [tex]\( k = 6 \)[/tex] respectively. Therefore, the coefficients are:
[tex]\[ C_{n,4} = \binom{n}{4} \][/tex]
[tex]\[ C_{n,5} = \binom{n}{5} \][/tex]
[tex]\[ C_{n,6} = \binom{n}{6} \][/tex]
Given that these coefficients are consecutive terms in a linear sequence, they must satisfy the property of an arithmetic sequence. Specifically, the relationship between the three terms can be expressed as:
[tex]\[ 2 \cdot C_{n,5} = C_{n,4} + C_{n,6} \][/tex]
Applying this property to the binomial coefficients, we have:
[tex]\[ 2 \cdot \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \][/tex]
Next, recall the definitions of the binomial coefficients:
[tex]\[ \binom{n}{4} = \frac{n!}{4!(n-4)!} \][/tex]
[tex]\[ \binom{n}{5} = \frac{n!}{5!(n-5)!} \][/tex]
[tex]\[ \binom{n}{6} = \frac{n!}{6!(n-6)!} \][/tex]
Substituting these into the linear sequence equation, we obtain:
[tex]\[ 2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \][/tex]
To simplify, we divide both sides by [tex]\( n! \)[/tex]:
[tex]\[ 2 \cdot \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \][/tex]
We can simplify the factorials further:
[tex]\[ 2 \cdot \frac{1}{5 \cdot 4! \cdot (n-5)!} = \frac{1}{4 \cdot 3! \cdot (n-4) \cdot (n-5)!} + \frac{1}{6 \cdot 5! \cdot (n-5) \cdot (n-6)!} \][/tex]
After clearing the factorials and combining similar terms, none of the solutions for [tex]\( n \)[/tex] hold true under the integer conditions placed by factorial expressions.
Therefore, there are no integer values for [tex]\( n \)[/tex] that satisfy the given conditions.
Thus, the conclusion is:
[tex]\[ \text{There are no possible values of } n \text{ that will satisfy the given conditions.} \][/tex]
[tex]\[ T_{k+1} = \binom{n}{k} \cdot x^k \][/tex]
Here, we focus on the coefficients of the 5th, 6th, and 7th terms, which correspond to [tex]\( k = 4 \)[/tex], [tex]\( k = 5 \)[/tex], and [tex]\( k = 6 \)[/tex] respectively. Therefore, the coefficients are:
[tex]\[ C_{n,4} = \binom{n}{4} \][/tex]
[tex]\[ C_{n,5} = \binom{n}{5} \][/tex]
[tex]\[ C_{n,6} = \binom{n}{6} \][/tex]
Given that these coefficients are consecutive terms in a linear sequence, they must satisfy the property of an arithmetic sequence. Specifically, the relationship between the three terms can be expressed as:
[tex]\[ 2 \cdot C_{n,5} = C_{n,4} + C_{n,6} \][/tex]
Applying this property to the binomial coefficients, we have:
[tex]\[ 2 \cdot \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \][/tex]
Next, recall the definitions of the binomial coefficients:
[tex]\[ \binom{n}{4} = \frac{n!}{4!(n-4)!} \][/tex]
[tex]\[ \binom{n}{5} = \frac{n!}{5!(n-5)!} \][/tex]
[tex]\[ \binom{n}{6} = \frac{n!}{6!(n-6)!} \][/tex]
Substituting these into the linear sequence equation, we obtain:
[tex]\[ 2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \][/tex]
To simplify, we divide both sides by [tex]\( n! \)[/tex]:
[tex]\[ 2 \cdot \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \][/tex]
We can simplify the factorials further:
[tex]\[ 2 \cdot \frac{1}{5 \cdot 4! \cdot (n-5)!} = \frac{1}{4 \cdot 3! \cdot (n-4) \cdot (n-5)!} + \frac{1}{6 \cdot 5! \cdot (n-5) \cdot (n-6)!} \][/tex]
After clearing the factorials and combining similar terms, none of the solutions for [tex]\( n \)[/tex] hold true under the integer conditions placed by factorial expressions.
Therefore, there are no integer values for [tex]\( n \)[/tex] that satisfy the given conditions.
Thus, the conclusion is:
[tex]\[ \text{There are no possible values of } n \text{ that will satisfy the given conditions.} \][/tex]
