If r is the remainder obtained on dividing `98^(5)` by 12, then the coefficient . of `x^(3)` in the binomial expansion of `(1+x/2)^(2r)` is:

Options

A. `55/2`
B. `91/2`
C. `102`
D. `70`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the remainder \( r \) when \( 98^5 \) is divided by 12.

To find \( r \), we first simplify \( 98 \) modulo \( 12 \):
\[ 98 \mod 12 = 2 \quad (\text{since } 98 = 8 \times 12 + 2) \]
Thus, we can rewrite \( 98^5 \) as:
\[ 98^5 \equiv 2^5 \mod 12 \]

Now, we calculate \( 2^5 \):
\[ 2^5 = 32 \]

Next, we find \( 32 \mod 12 \):
\[ 32 \div 12 = 2 \quad \text{(which gives a quotient of 2)} \]
\[ 32 - (12 \times 2) = 32 - 24 = 8 \]
Thus, the remainder \( r \) is:
\[ r = 8 \]

### Step 2: Find the coefficient of \( x^3 \) in the binomial expansion of \( \left(1 + \frac{x}{2}\right)^{2r} \).

Since we found \( r = 8 \), we need to evaluate:
\[ \left(1 + \frac{x}{2}\right)^{2 \times 8} = \left(1 + \frac{x}{2}\right)^{16} \]

### Step 3: Use the binomial theorem to find the coefficient of \( x^3 \).

The binomial expansion states:
\[ (1 + y)^n = \sum_{k=0}^{n} \binom{n}{k} y^k \]
In our case, \( y = \frac{x}{2} \) and \( n = 16 \). We want the coefficient of \( x^3 \), which corresponds to \( k = 3 \):
\[ \text{Coefficient of } x^3 = \binom{16}{3} \left(\frac{1}{2}\right)^3 \]

### Step 4: Calculate \( \binom{16}{3} \).

Using the formula for combinations:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
We have:
\[ \binom{16}{3} = \frac{16!}{3!(16-3)!} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = \frac{3360}{6} = 560 \]

### Step 5: Calculate the coefficient of \( x^3 \).

Now substituting back:
\[ \text{Coefficient of } x^3 = 560 \times \left(\frac{1}{2}\right)^3 = 560 \times \frac{1}{8} = 70 \]

### Final Answer:
Thus, the coefficient of \( x^3 \) in the expansion of \( \left(1 + \frac{x}{2}\right)^{16} \) is:
\[ \boxed{70} \]