Understanding Binomial Expansion and Arithmetic Progression

The question asks for the value of a positive integer $n$, given that the coefficients of the 2nd, 3rd, and 4th terms in the binomial expansion of $(1+x)^n$ are in an arithmetic progression (A.P.).

Binomial Expansion of $(1+x)^n$

The binomial expansion of $(1+x)^n$ is given by the formula:

$(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots + \binom{n}{n}x^n$

The $(r+1)^{th}$ term in this expansion is $T_{r+1} = \binom{n}{r} x^r$, and its coefficient is $\binom{n}{r}$.

Identifying the Coefficients

We need the coefficients of the 2nd, 3rd, and 4th terms:

• The 2nd term is $T_2$. Here, $r+1 = 2$, so $r=1$. The coefficient is $\binom{n}{1}$.
• The 3rd term is $T_3$. Here, $r+1 = 3$, so $r=2$. The coefficient is $\binom{n}{2}$.
• The 4th term is $T_4$. Here, $r+1 = 4$, so $r=3$. The coefficient is $\binom{n}{3}$.

These coefficients are $\binom{n}{1}$, $\binom{n}{2}$, and $\binom{n}{3}$.

Applying the Arithmetic Progression Condition

If three numbers $a, b, c$ are in A.P., then the middle term is the average of the other two, i.e., $b = \frac{a+c}{2}$, which can be rearranged as $2b = a+c$.

In this case, the coefficients $\binom{n}{1}$, $\binom{n}{2}$, and $\binom{n}{3}$ are in A.P. So, we have:

$2 \binom{n}{2} = \binom{n}{1} + \binom{n}{3}$

Expressing Binomial Coefficients using Factorials

Recall the formula for binomial coefficients: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Let's express the required coefficients:

• $\binom{n}{1} = \frac{n!}{1!(n-1)!} = \frac{n \times (n-1)!}{1 \times (n-1)!} = n$
• $\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)(n-2)!}{2 \times 1 \times (n-2)!} = \frac{n(n-1)}{2}$
• $\binom{n}{3} = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)(n-3)!}{3 \times 2 \times 1 \times (n-3)!} = \frac{n(n-1)(n-2)}{6}$

For the 4th term to exist with a non-zero coefficient in this context, $n$ must be at least 3 (i.e., $n \ge 3$).

Setting up and Solving the Equation for n

Substitute the expressions for the coefficients into the A.P. equation:

$2 \left( \frac{n(n-1)}{2} \right) = n + \frac{n(n-1)(n-2)}{6}$

Simplify the left side:

$n(n-1) = n + \frac{n(n-1)(n-2)}{6}$

Since $n \ge 3$, we know $n \ne 0$. We can divide both sides of the equation by $n$:

$n-1 = 1 + \frac{(n-1)(n-2)}{6}$

Subtract 1 from both sides:

$n-1-1 = \frac{(n-1)(n-2)}{6}$

$n-2 = \frac{(n-1)(n-2)}{6}$

Rearrange the equation to set it to zero:

$n-2 - \frac{(n-1)(n-2)}{6} = 0$

Factor out the common term $(n-2)$:

$(n-2) \left( 1 - \frac{n-1}{6} \right) = 0$

Simplify the term in the parenthesis:

$(n-2) \left( \frac{6 - (n-1)}{6} \right) = 0$

$(n-2) \left( \frac{6 - n + 1}{6} \right) = 0$

$(n-2) \left( \frac{7 - n}{6} \right) = 0$

This equation gives two possible solutions:

• Case 1: $n-2 = 0 \implies n = 2$
• Case 2: $\frac{7-n}{6} = 0 \implies 7-n = 0 \implies n = 7$

Checking the Validity of Solutions

We established earlier that for the coefficient $\binom{n}{3}$ to be well-defined and for the 4th term to exist in the sequence of terms being considered for A.P., $n$ must be a positive integer and $n \ge 3$.

• If $n=2$, the expansion $(1+x)^2$ has only 3 terms ($\binom{2}{0}, \binom{2}{1}, \binom{2}{2}$). The 4th term does not exist. Therefore, $n=2$ is not a valid solution in this context.
• If $n=7$, $n$ is a positive integer and $n \ge 3$. Let's check if the coefficients are in A.P.:
  • Coefficient of 2nd term: $\binom{7}{1} = 7$
  • Coefficient of 3rd term: $\binom{7}{2} = \frac{7 \times 6}{2} = 21$
  • Coefficient of 4th term: $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
  The coefficients are 7, 21, 35. Check if they form an A.P.: $21 - 7 = 14$ and $35 - 21 = 14$. The common difference is 14. Thus, 7, 21, and 35 are in A.P. So, $n=7$ is a valid solution.

Therefore, the value of $n$ is 7.

Revision Table: Binomial Coefficient A.P. Problem

Additional Information: Properties of Binomial Expansion Coefficients

The coefficients in a binomial expansion $(1+x)^n$, given by $\binom{n}{r}$ for $r=0, 1, \dots, n$, exhibit several interesting properties:

• Symmetry: $\binom{n}{r} = \binom{n}{n-r}$. The coefficients are symmetric about the middle term(s).
• Sum of Coefficients: The sum of all coefficients in $(1+x)^n$ is $2^n$. This is found by setting $x=1$: $(1+1)^n = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n}$.
• Maximum Coefficient: The largest coefficient occurs at the middle term(s). If $n$ is even, the maximum is $\binom{n}{n/2}$. If $n$ is odd, there are two equal maximum coefficients: $\binom{n}{(n-1)/2}$ and $\binom{n}{(n+1)/2}$.
• Relationship between consecutive coefficients: $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$. This identity is the basis for Pascal's Triangle.
• Behavior of coefficients: For small values of $r$, the coefficients $\binom{n}{r}$ generally increase as $r$ increases, up to the middle term(s), and then decrease. The fact that $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}$ can be in A.P. imposes a specific condition on $n$.

Problems involving binomial coefficients in A.P., G.P., or H.P. often require setting up equations using the definitions of these progressions and solving for $n$ or other variables.