If `S_(n)` denotes the sum fo n terms of a G.P. whose first term and common ratio are a and r respectively then show that: `S_(1)+S_(3)+S_(5)+……..+S_(2n)-1=(an)/(1-r)-(ar(1-r^(2n)))/((1-r^(2))`

Answer

To solve the problem, we need to show that:

\[ S_1 + S_3 + S_5 + \ldots + S_{2n-1} = \frac{an}{1-r} - \frac{ar(1 - r^{2n})}{1 - r^2} \]

where \( S_n \) is the sum of the first \( n \) terms of a geometric progression (G.P.) with first term \( a \) and common ratio \( r \).

### Step-by-step Solution:

1. **Recall the formula for the sum of the first \( n \) terms of a G.P.:**
\[ S_n = \frac{a(1 - r^n)}{1 - r} \quad \text{(for } r \neq 1\text{)} \]

2. **Calculate \( S_1, S_3, S_5, \ldots, S_{2n-1} \):**
- For \( S_1 \):
\[ S_1 = \frac{a(1 - r^1)}{1 - r} = \frac{a(1 - r)}{1 - r} = a \]
- For \( S_3 \):
\[ S_3 = \frac{a(1 - r^3)}{1 - r} \]
- For \( S_5 \):
\[ S_5 = \frac{a(1 - r^5)}{1 - r} \]
- Continuing this way, we find:
\[ S_{2n-1} = \frac{a(1 - r^{2n-1})}{1 - r} \]

3. **Sum \( S_1 + S_3 + S_5 + \ldots + S_{2n-1} \):**
\[ S_1 + S_3 + S_5 + \ldots + S_{2n-1} = \frac{a}{1 - r} + \frac{a(1 - r^3)}{1 - r} + \frac{a(1 - r^5)}{1 - r} + \ldots + \frac{a(1 - r^{2n-1})}{1 - r} \]
Factoring out \( \frac{a}{1 - r} \):
\[ = \frac{a}{1 - r} \left( 1 + (1 - r^3) + (1 - r^5) + \ldots + (1 - r^{2n-1}) \right) \]

4. **Count the number of terms:**
There are \( n \) terms in the series \( S_1, S_3, S_5, \ldots, S_{2n-1} \).

5. **Simplify the series inside the parentheses:**
The series can be rewritten as:
\[ n - (r + r^3 + r^5 + \ldots + r^{2n-1}) \]
The series \( r + r^3 + r^5 + \ldots + r^{2n-1} \) is a G.P. with first term \( r \) and common ratio \( r^2 \).

6. **Sum the G.P. \( r + r^3 + r^5 + \ldots + r^{2n-1} \):**
The number of terms in this series is \( n \):
\[ \text{Sum} = \frac{r(1 - r^{2n})}{1 - r^2} \]

7. **Substituting back:**
\[ S_1 + S_3 + S_5 + \ldots + S_{2n-1} = \frac{a}{1 - r} \left( n - \frac{r(1 - r^{2n})}{1 - r^2} \right) \]

8. **Final simplification:**
\[ = \frac{an}{1 - r} - \frac{ar(1 - r^{2n})}{(1 - r)(1 - r^2)} \]

9. **Rearranging gives us the desired result:**
\[ = \frac{an}{1 - r} - \frac{ar(1 - r^{2n})}{1 - r^2} \]

### Conclusion:
Thus, we have shown that:

\[ S_1 + S_3 + S_5 + \ldots + S_{2n-1} = \frac{an}{1 - r} - \frac{ar(1 - r^{2n})}{1 - r^2} \]