`3""^(n)C_(0)+10""^(n)C_(1)+28""^(n)C_(2)+82""^(n)C_(3)+…(n+1)` terms =

Options

A. `n(14n-4)+3`
B. `2^(n+2)+4n+14n(n-1)-1`
C. `3^(n-1)(4n+9)`
D. `3^(n+1)(n-1)2^(n)`

Answer

To solve the problem, we need to find the sum of the series given by:

\[ S = 3 \cdot \binom{n}{0} + 10 \cdot \binom{n}{1} + 28 \cdot \binom{n}{2} + 82 \cdot \binom{n}{3} + \ldots + (n+1) \text{ terms} \]

### Step-by-Step Solution:

1. **Identify the Pattern**:
The coefficients (3, 10, 28, 82, ...) suggest a pattern. We can express the general term as \( a_r = 3^{r+1} \) for \( r = 0, 1, 2, 3, \ldots, n \).

2. **Express the Sum**:
The sum can be rewritten using summation notation:
\[ S = \sum_{r=0}^{n} a_r \cdot \binom{n}{r} \]
where \( a_r = 3^{r+1} \).

3. **Factor out constants**:
We can factor out \( 3 \) from \( a_r \):
\[ S = 3 \sum_{r=0}^{n} 3^r \cdot \binom{n}{r} \]

4. **Use Binomial Theorem**:
According to the binomial theorem, we know:
\[ \sum_{r=0}^{n} x^r \cdot \binom{n}{r} = (1 + x)^n \]
Setting \( x = 3 \):
\[ \sum_{r=0}^{n} 3^r \cdot \binom{n}{r} = (1 + 3)^n = 4^n \]

5. **Substitute Back**:
Now, substituting back into our expression for \( S \):
\[ S = 3 \cdot 4^n \]

6. **Final Result**:
Therefore, the sum of the series is:
\[ S = 3 \cdot 4^n \]