If `sum_(r=1)^n I(r)=(3^n -1)` , then `sum_(r=1)^n 1/(I(r))` is equal to :

Options

A. `2(1-(1/3)^n)`
B. `(1-(1/3)^n)`
C. `3/4(1-(1/3)^n)`
D. `4/3(1-(1/3)^n)`

Answer

To solve the problem, we need to find the value of the summation \( \sum_{r=1}^{n} \frac{1}{I(r)} \) given that \( \sum_{r=1}^{n} I(r) = 3^n - 1 \).

### Step-by-step Solution:

1. **Understanding the Given Summation**:
We know that:
\[ \sum_{r=1}^{n} I(r) = 3^n - 1 \]
This implies that the series \( I(r) \) can be expressed in terms of a geometric progression (GP).

2. **Identifying the Terms of the Series**:
We can assume that \( I(r) \) is of the form:
\[ I(r) = 2 \cdot 3^{r-1} \]
This is because the sum of a GP with first term \( a = 1 \) and common ratio \( r = 3 \) gives us:
\[ S_n = a \frac{r^n - 1}{r - 1} = \frac{3^n - 1}{2} \]
Multiplying through by 2 gives us \( 3^n - 1 \).

3. **Calculating \( \sum_{r=1}^{n} \frac{1}{I(r)} \)**:
Now, we need to compute:
\[ \sum_{r=1}^{n} \frac{1}{I(r)} = \sum_{r=1}^{n} \frac{1}{2 \cdot 3^{r-1}} = \frac{1}{2} \sum_{r=1}^{n} \frac{1}{3^{r-1}} \]
The series \( \sum_{r=1}^{n} \frac{1}{3^{r-1}} \) is also a geometric series with first term \( 1 \) and common ratio \( \frac{1}{3} \).

4. **Using the Formula for the Sum of a GP**:
The sum of the first \( n \) terms of a GP is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Here, \( a = 1 \) and \( r = \frac{1}{3} \):
\[ \sum_{r=1}^{n} \frac{1}{3^{r-1}} = \frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{3^n}}{\frac{2}{3}} = \frac{3}{2} \left(1 - \frac{1}{3^n}\right) \]

5. **Substituting Back**:
Now substituting back into our equation:
\[ \sum_{r=1}^{n} \frac{1}{I(r)} = \frac{1}{2} \cdot \frac{3}{2} \left(1 - \frac{1}{3^n}\right) = \frac{3}{4} \left(1 - \frac{1}{3^n}\right) \]

6. **Final Result**:
Therefore, the final result is:
\[ \sum_{r=1}^{n} \frac{1}{I(r)} = \frac{3}{4} \left(1 - \frac{1}{3^n}\right) \]