If the coefficients of `(p+1)`th and `(P+3)`th terms in the expansion of `(1+x)^(2n)` are equal then prove that n=p+1

Options

A. p = n-2
B. p= n-1
C. p=n+1
D. p=2n-2

Answer

To solve the problem, we need to show that if the coefficients of the \((p+1)\)th and \((p+3)\)th terms in the expansion of \((1+x)^{2n}\) are equal, then \(n = p + 1\).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the binomial expansion of \((1+x)^{2n}\) is given by:
\[ T_r = \binom{2n}{r} x^r \]
where \(T_r\) is the \((r+1)\)th term.

2. **Coefficients of the Relevant Terms**:
- The coefficient of the \((p+1)\)th term (which corresponds to \(r = p\)) is:
\[ \text{Coefficient of } T_{p} = \binom{2n}{p} \]
- The coefficient of the \((p+3)\)th term (which corresponds to \(r = p + 2\)) is:
\[ \text{Coefficient of } T_{p+2} = \binom{2n}{p + 2} \]

3. **Set the Coefficients Equal**:
According to the problem, these coefficients are equal:
\[ \binom{2n}{p} = \binom{2n}{p + 2} \]

4. **Using the Property of Binomial Coefficients**:
The property of binomial coefficients states that:
\[ \binom{n}{k} = \binom{n}{n-k} \]
This implies that if \(\binom{n}{k_1} = \binom{n}{k_2}\), then either:
- \(k_1 = k_2\) or
- \(k_1 + k_2 = n\)

In our case, we have:
- \(k_1 = p\)
- \(k_2 = p + 2\)
- \(n = 2n\)

Therefore, we can set up the equation:
\[ p + (p + 2) = 2n \]

5. **Simplify the Equation**:
Simplifying the equation gives:
\[ 2p + 2 = 2n \]
Dividing through by 2:
\[ p + 1 = n \]

6. **Conclusion**:
Thus, we have shown that:
\[ n = p + 1 \]