Find the sum of `2n` terms of the series whose every even term is `' a '` times the term before it and every odd term is `' c '` times the term before it, the first term being unity.

Explanation

To find the sum of `2n` terms of the series where every even term is `'a'` times the term before it and every odd term is `'c'` times the term before it, starting with the first term as unity, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Series Structure**:
- The first term \( T_1 = 1 \).
- The second term \( T_2 = a \times T_1 = a \).
- The third term \( T_3 = c \times T_2 = ac \).
- The fourth term \( T_4 = a \times T_3 = a^2c \).
- The fifth term \( T_5 = c \times T_4 = a^2c^2 \).
- Continuing this pattern, we can see that the series alternates between multiplying by \( a \) and \( c \).

2. **Separate Odd and Even Terms**:
- The odd terms are \( T_1, T_3, T_5, \ldots \) which can be expressed as:
\[ 1, ac, a^2c^2, a^3c^3, \ldots \]
- The even terms are \( T_2, T_4, T_6, \ldots \) which can be expressed as:
\[ a, a^2c, a^3c^2, \ldots \]

3. **Sum of Odd Terms**:
- The odd terms form a geometric progression (GP) with:
- First term \( a_1 = 1 \)
- Common ratio \( r = ac \)
- Number of terms \( n \)
- The sum of the first \( n \) terms of a GP is given by:
\[ S_n = \frac{a_1 (r^n - 1)}{r - 1} \]
- Therefore, the sum of the odd terms is:
\[ S_{odd} = \frac{1 \cdot ((ac)^n - 1)}{ac - 1} = \frac{(ac)^n - 1}{ac - 1} \]

4. **Sum of Even Terms**:
- The even terms also form a GP with:
- First term \( a_1 = a \)
- Common ratio \( r = ac \)
- Number of terms \( n \)
- Thus, the sum of the even terms is:
\[ S_{even} = \frac{a \cdot ((ac)^n - 1)}{ac - 1} \]

5. **Total Sum of the Series**:
- The total sum \( S \) of the series is the sum of the odd terms and the even terms:
\[ S = S_{odd} + S_{even} \]
- Substituting the sums we found:
\[ S = \frac{(ac)^n - 1}{ac - 1} + \frac{a \cdot ((ac)^n - 1)}{ac - 1} \]
- Factoring out the common denominator:
\[ S = \frac{(ac)^n - 1 + a((ac)^n - 1)}{ac - 1} \]
- Simplifying:
\[ S = \frac{(ac)^n - 1 + a(ac)^n - a}{ac - 1} = \frac{(1 + a)(ac)^n - (1 + a)}{ac - 1} \]
- Therefore, the final answer is:
\[ S = \frac{(1 + a)((ac)^n - 1)}{ac - 1} \]