Find the sum to
n terms of `3 (3)/(8) + 2 (1)/(4) + 1 (1)/(2)+ .....`

Answer

To find the sum to \( n \) terms of the series \( 3 \frac{3}{8} + 2 \frac{1}{4} + 1 \frac{1}{2} + \ldots \), we can follow these steps:

### Step 1: Write the terms in proper fraction form
The first three terms can be converted to improper fractions:
- \( 3 \frac{3}{8} = \frac{27}{8} \)
- \( 2 \frac{1}{4} = \frac{9}{4} \)
- \( 1 \frac{1}{2} = \frac{3}{2} \)

So, the series can be expressed as:
\[ S = \frac{27}{8} + \frac{9}{4} + \frac{3}{2} + \ldots \]

### Step 2: Identify the pattern and common ratio
To determine if this series is a geometric progression (GP), we can find the ratio of consecutive terms:
- The second term divided by the first term:
\[ \frac{\frac{9}{4}}{\frac{27}{8}} = \frac{9}{4} \times \frac{8}{27} = \frac{72}{108} = \frac{2}{3} \]
- The third term divided by the second term:
\[ \frac{\frac{3}{2}}{\frac{9}{4}} = \frac{3}{2} \times \frac{4}{9} = \frac{12}{18} = \frac{2}{3} \]

Since the ratio is constant, we confirm that this is a GP with a common ratio \( r = \frac{2}{3} \).

### Step 3: Identify the first term
The first term \( a \) of the GP is:
\[ a = \frac{27}{8} \]

### Step 4: Use the formula for the sum of the first \( n \) terms of a GP
The formula for the sum of the first \( n \) terms of a GP is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Substituting the values of \( a \) and \( r \):
\[ S_n = \frac{27}{8} \cdot \frac{1 - \left(\frac{2}{3}\right)^n}{1 - \frac{2}{3}} \]

### Step 5: Simplify the expression
The denominator simplifies to:
\[ 1 - \frac{2}{3} = \frac{1}{3} \]
Thus, we can rewrite the sum:
\[ S_n = \frac{27}{8} \cdot \frac{1 - \left(\frac{2}{3}\right)^n}{\frac{1}{3}} = \frac{27}{8} \cdot 3 \cdot \left(1 - \left(\frac{2}{3}\right)^n\right) \]
This simplifies to:
\[ S_n = \frac{81}{8} \left(1 - \left(\frac{2}{3}\right)^n\right) \]

### Final Result
The sum to \( n \) terms of the series is:
\[ S_n = \frac{81}{8} \left(1 - \left(\frac{2}{3}\right)^n\right) \]