Sum of n terms:
(i) `5+55+555+`…………..
(ii) `9+99+999+`…………..
(iii) `3+33+333+`………………
(v) `8+88+888+`…………….

Answer

To find the sum of the first n terms of the given series, we will analyze each part step by step.

### Part (i): Sum of the series \(5 + 55 + 555 + \ldots\)

1. **Identify the pattern**: The terms can be expressed as:
- \(5 = 5\)
- \(55 = 5 \times 11\)
- \(555 = 5 \times 111\)
- Continuing this, we can express the nth term \(T_n\) as:
\[ T_n = 5 \times (10^0 + 10^1 + 10^2 + \ldots + 10^{n-1}) \]

2. **Sum of the geometric series**: The sum of the geometric series \(10^0 + 10^1 + 10^2 + \ldots + 10^{n-1}\) can be calculated using the formula:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
where \(a = 1\) (the first term), \(r = 10\) (the common ratio), and \(n\) is the number of terms.
\[ S_n = \frac{1(10^n - 1)}{10 - 1} = \frac{10^n - 1}{9} \]

3. **Substituting back**: Therefore, the sum of the first n terms becomes:
\[ S_n = 5 \times \frac{10^n - 1}{9} = \frac{5(10^n - 1)}{9} \]

### Part (ii): Sum of the series \(9 + 99 + 999 + \ldots\)

1. **Identify the pattern**: The terms can be expressed as:
- \(9 = 9\)
- \(99 = 9 \times 11\)
- \(999 = 9 \times 111\)
- The nth term \(T_n\) can be expressed as:
\[ T_n = 9 \times (10^0 + 10^1 + 10^2 + \ldots + 10^{n-1}) \]

2. **Using the sum of the geometric series**:
\[ S_n = \frac{10^n - 1}{9} \]

3. **Substituting back**:
\[ S_n = 9 \times \frac{10^n - 1}{9} = 10^n - 1 \]

### Part (iii): Sum of the series \(3 + 33 + 333 + \ldots\)

1. **Identify the pattern**: The terms can be expressed as:
- \(3 = 3\)
- \(33 = 3 \times 11\)
- \(333 = 3 \times 111\)
- The nth term \(T_n\) can be expressed as:
\[ T_n = 3 \times (10^0 + 10^1 + 10^2 + \ldots + 10^{n-1}) \]

2. **Using the sum of the geometric series**:
\[ S_n = \frac{10^n - 1}{9} \]

3. **Substituting back**:
\[ S_n = 3 \times \frac{10^n - 1}{9} = \frac{3(10^n - 1)}{9} = \frac{10^n - 1}{3} \]

### Part (iv): Sum of the series \(8 + 88 + 888 + \ldots\)

1. **Identify the pattern**: The terms can be expressed as:
- \(8 = 8\)
- \(88 = 8 \times 11\)
- \(888 = 8 \times 111\)
- The nth term \(T_n\) can be expressed as:
\[ T_n = 8 \times (10^0 + 10^1 + 10^2 + \ldots + 10^{n-1}) \]

2. **Using the sum of the geometric series**:
\[ S_n = \frac{10^n - 1}{9} \]

3. **Substituting back**:
\[ S_n = 8 \times \frac{10^n - 1}{9} = \frac{8(10^n - 1)}{9} \]

### Final Answers:
- (i) \(S_n = \frac{5(10^n - 1)}{9}\)
- (ii) \(S_n = 10^n - 1\)
- (iii) \(S_n = \frac{10^n - 1}{3}\)
- (iv) \(S_n = \frac{8(10^n - 1)}{9}\)