If `.^nC_0+3.^nC_1+5^nC_2+7^nC_3+ . . .till (n+1) term=2^100*101` then the value of `2[(n-1)/2]` where `[.]` is G.I.F)

Options

A. 98
B. 97
C. 96
D. 100

Answer

To solve the problem, we need to evaluate the expression given and find the value of \(2 \left\lfloor \frac{n-1}{2} \right\rfloor\).

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression given is:
\[ \sum_{r=0}^{n} (2r + 1) \binom{n}{r} \]
This can be rewritten as:
\[ \sum_{r=0}^{n} 2r \binom{n}{r} + \sum_{r=0}^{n} \binom{n}{r} \]

2. **Using Binomial Theorems**:
From the binomial theorem, we know:
\[ \sum_{r=0}^{n} \binom{n}{r} = 2^n \]
For the first part, we can use the identity:
\[ \sum_{r=0}^{n} r \binom{n}{r} = n \cdot 2^{n-1} \]
Therefore, we have:
\[ \sum_{r=0}^{n} 2r \binom{n}{r} = 2n \cdot 2^{n-1} = n \cdot 2^n \]

3. **Combining the Results**:
Now substituting back into our expression:
\[ n \cdot 2^n + 2^n = (n + 1) \cdot 2^n \]

4. **Setting the Equation**:
We are given that:
\[ (n + 1) \cdot 2^n = 2^{100} \cdot 101 \]
Dividing both sides by \(2^n\):
\[ n + 1 = 101 \cdot 2^{100 - n} \]

5. **Finding \(n\)**:
To find \(n\), we can analyze the equation. Since \(n + 1\) must be an integer, \(2^{100 - n}\) must also yield an integer. This implies that \(100 - n\) must be non-negative, hence \(n \leq 100\).

6. **Finding \(n\)**:
We can set \(n + 1 = 101\) when \(n = 100\):
\[ 100 + 1 = 101 \cdot 2^{0} \implies n = 100 \]

7. **Calculating \(2 \left\lfloor \frac{n-1}{2} \right\rfloor\)**:
Now we need to find:
\[ 2 \left\lfloor \frac{100 - 1}{2} \right\rfloor = 2 \left\lfloor \frac{99}{2} \right\rfloor = 2 \cdot 49 = 98 \]

### Final Answer:
Thus, the value of \(2 \left\lfloor \frac{n-1}{2} \right\rfloor\) is \(98\).