Suppose a series of `n` terms given by `S_(n)=t_(1)+t_(2)+t_(3)+`. . ..`+t_(n)`
then `S_(n-1)=t_(1)+t_(2)+t_(3)+` . . . `+t_(n-1),nge1` subtracting we get `S_(n)-S_(n-1)=t_(n),nge2` surther if we put `n=1` is the first sum then `S_(1)=t_(1)` thus w can write `t_(n)=S_(n)-S_(n-1),nge2` and `t_(1)=S_(1)`
Q. if the sum of `n` terms of a series `a.2^(n)-b` then the sum `sum_(r=1)^(infty)(1)/(t_(r))` is

Options

A. `a`
B. `(a)/(2)`
C. `(2)/(a)`
D. `(1)/(a)`

Explanation

To solve the problem step by step, we start with the given information and derive the required summation.

### Step 1: Understand the given series
We are given that the sum of the first `n` terms of a series is given by:
\[ S_n = t_1 + t_2 + t_3 + \ldots + t_n \]
and for \( n \geq 1 \):
\[ S_{n-1} = t_1 + t_2 + t_3 + \ldots + t_{n-1} \]

### Step 2: Find the relationship between \( S_n \) and \( S_{n-1} \)
By subtracting \( S_{n-1} \) from \( S_n \):
\[ S_n - S_{n-1} = t_n \]
This holds for \( n \geq 2 \).

### Step 3: Establish the expression for \( t_n \)
From the above relationship, we can express \( t_n \) as:
\[ t_n = S_n - S_{n-1} \]
For \( n = 1 \):
\[ S_1 = t_1 \]

### Step 4: Given the sum of the series
It is given that the sum of the first \( n \) terms of the series is:
\[ S_n = a \cdot 2^n - b \]

### Step 5: Calculate \( S_1 \)
For \( n = 1 \):
\[ S_1 = a \cdot 2^1 - b = 2a - b \]
Thus, we have:
\[ t_1 = 2a - b \]

### Step 6: Calculate \( t_n \)
Using the expression for \( S_n \):
\[ S_n = a \cdot 2^n - b \]
\[ S_{n-1} = a \cdot 2^{n-1} - b \]
Now substituting these into the expression for \( t_n \):
\[ t_n = S_n - S_{n-1} = (a \cdot 2^n - b) - (a \cdot 2^{n-1} - b) \]
This simplifies to:
\[ t_n = a \cdot 2^n - a \cdot 2^{n-1} = a \cdot (2^n - 2^{n-1}) = a \cdot 2^{n-1} \]

### Step 7: Find the summation \( \sum_{r=1}^{\infty} \frac{1}{t_r} \)
We know:
\[ t_r = a \cdot 2^{r-1} \]
Thus:
\[ \sum_{r=1}^{\infty} \frac{1}{t_r} = \sum_{r=1}^{\infty} \frac{1}{a \cdot 2^{r-1}} = \frac{1}{a} \sum_{r=1}^{\infty} \frac{1}{2^{r-1}} \]

### Step 8: Evaluate the infinite series
The series \( \sum_{r=1}^{\infty} \frac{1}{2^{r-1}} \) is a geometric series with the first term \( 1 \) and common ratio \( \frac{1}{2} \):
\[ \sum_{r=1}^{\infty} \frac{1}{2^{r-1}} = \frac{1}{1 - \frac{1}{2}} = 2 \]

### Step 9: Final result
Thus:
\[ \sum_{r=1}^{\infty} \frac{1}{t_r} = \frac{1}{a} \cdot 2 = \frac{2}{a} \]

### Final Answer:
\[ \sum_{r=1}^{\infty} \frac{1}{t_r} = \frac{2}{a} \]

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