If `n^(th)` term `(T_(n))` of the sequece is given by the relation `T_(n+1)=T_(n)+2^(n),(n epsilonN)` and `T_(1)=3`, then the value of `sum_(r=1)^(oo)1/(T_(r)-1)` equals

Options

A. `1`
B. `2`
C. `3`
D. `4`

Answer

To solve the problem, we need to find the value of the infinite sum \( \sum_{r=1}^{\infty} \frac{1}{T_r - 1} \), given the recurrence relation for the sequence \( T_n \) and the initial condition \( T_1 = 3 \).

### Step-by-Step Solution:

1. **Understand the recurrence relation**:
We are given that \( T_{n+1} = T_n + 2^n \) and \( T_1 = 3 \).

2. **Calculate the first few terms of the sequence**:
- For \( n = 1 \):
\[ T_2 = T_1 + 2^1 = 3 + 2 = 5 \]
- For \( n = 2 \):
\[ T_3 = T_2 + 2^2 = 5 + 4 = 9 \]
- For \( n = 3 \):
\[ T_4 = T_3 + 2^3 = 9 + 8 = 17 \]
- For \( n = 4 \):
\[ T_5 = T_4 + 2^4 = 17 + 16 = 33 \]

So, the first few terms are:
- \( T_1 = 3 \)
- \( T_2 = 5 \)
- \( T_3 = 9 \)
- \( T_4 = 17 \)
- \( T_5 = 33 \)

3. **Express \( T_r - 1 \)**:
- We can compute \( T_r - 1 \) for the first few terms:
- \( T_1 - 1 = 3 - 1 = 2 \)
- \( T_2 - 1 = 5 - 1 = 4 \)
- \( T_3 - 1 = 9 - 1 = 8 \)
- \( T_4 - 1 = 17 - 1 = 16 \)
- \( T_5 - 1 = 33 - 1 = 32 \)

4. **Identify the pattern**:
- Notice that:
- \( T_1 - 1 = 2 = 2^1 \)
- \( T_2 - 1 = 4 = 2^2 \)
- \( T_3 - 1 = 8 = 2^3 \)
- \( T_4 - 1 = 16 = 2^4 \)
- \( T_5 - 1 = 32 = 2^5 \)
- This suggests that \( T_r - 1 = 2^r \).

5. **Substitute into the sum**:
- Now we can rewrite the sum:
\[ \sum_{r=1}^{\infty} \frac{1}{T_r - 1} = \sum_{r=1}^{\infty} \frac{1}{2^r} \]

6. **Calculate the infinite geometric series**:
- The series \( \sum_{r=1}^{\infty} \frac{1}{2^r} \) is a geometric series with first term \( a = \frac{1}{2} \) and common ratio \( r = \frac{1}{2} \).
- The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
- Substituting the values:
\[ S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]

### Final Answer:
The value of \( \sum_{r=1}^{\infty} \frac{1}{T_r - 1} \) is \( 1 \).

---