The value of `m ,` for which the coefficients of the `(2m+1)` th terms in the expansion of `(1+x)^(10)` are equal is `3` b. `1` c. `5` d. 8

Options

A. 3
B. 1
C. 5
D. 8

Answer

To find the value of \( m \) for which the coefficients of the \( (2m+1) \)th term and the \( (4m+5) \)th term in the expansion of \( (1+x)^{10} \) are equal, we can follow these steps:

### Step 1: Write the general term in the expansion
The \( r \)th term in the expansion of \( (1+x)^{n} \) is given by:
\[ T_{r+1} = \binom{n}{r} x^r \]
For \( n = 10 \), the \( r \)th term becomes:
\[ T_{r+1} = \binom{10}{r} x^r \]

### Step 2: Identify the terms we are interested in
We need to find the coefficients of the \( (2m+1) \)th term and the \( (4m+5) \)th term.

- The coefficient of the \( (2m+1) \)th term is:
\[ \text{Coefficient of } T_{2m+1} = \binom{10}{2m} \]

- The coefficient of the \( (4m+5) \)th term is:
\[ \text{Coefficient of } T_{4m+5} = \binom{10}{4m+4} \]

### Step 3: Set the coefficients equal
We set the coefficients equal to each other:
\[ \binom{10}{2m} = \binom{10}{4m+4} \]

### Step 4: Use properties of binomial coefficients
From the property of binomial coefficients, we know:
\[ \binom{n}{k} = \binom{n}{n-k} \]
This gives us two cases to consider:

#### Case 1: \( 2m = 4m + 4 \)
Solving this equation:
\[ 2m - 4m = 4 \implies -2m = 4 \implies m = -2 \]
Since \( m = -2 \) is not a valid option (as it would lead to a negative term), we discard this case.

#### Case 2: \( 2m + (4m + 4) = 10 \)
Solving this equation:
\[ 2m + 4m + 4 = 10 \implies 6m + 4 = 10 \implies 6m = 6 \implies m = 1 \]

### Conclusion
The value of \( m \) for which the coefficients of the \( (2m+1) \)th term and the \( (4m+5) \)th term are equal is:
\[ \boxed{1} \]