If `X = (a)/((1+r)) + (a)/((1+r)^2) + ….+ (a)/((1+r)^n)`, then what is the value of a + a (1+ r) + ... + a `(1 + r)^(n–1)` ?

Options

A. ` X [(1 + r) + (1+ r)^2 + ... + (1 + r)_n] `
B. ` X (1 + r)^n`
C. `X [(1 + r)^n – 1//r]
D. `X (1 + r)^(n–1)`

Answer

To solve the problem, we need to find the value of \( a + a(1 + r) + a(1 + r)^2 + \ldots + a(1 + r)^{n-1} \) given that

\[ X = \frac{a}{1 + r} + \frac{a}{(1 + r)^2} + \ldots + \frac{a}{(1 + r)^n}. \]

### Step 1: Recognize the series

The expression for \( X \) is a geometric series where the first term is \( \frac{a}{1 + r} \) and the common ratio is \( \frac{1}{1 + r} \). The number of terms in this series is \( n \).

### Step 2: Use the formula for the sum of a geometric series

The sum \( S \) of a geometric series can be calculated using the formula:

\[ S_n = a \frac{1 - r^n}{1 - r}, \]

where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

In our case:

- The first term \( a_1 = \frac{a}{1 + r} \)
- The common ratio \( r = \frac{1}{1 + r} \)
- The number of terms \( n \)

Thus, we can write:

\[ X = \frac{a}{1 + r} \cdot \frac{1 - \left(\frac{1}{1 + r}\right)^n}{1 - \frac{1}{1 + r}}. \]

### Step 3: Simplify the expression for \( X \)

The denominator simplifies to:

\[ 1 - \frac{1}{1 + r} = \frac{(1 + r) - 1}{1 + r} = \frac{r}{1 + r}. \]

Thus, we have:

\[ X = \frac{a}{1 + r} \cdot \frac{1 - \left(\frac{1}{1 + r}\right)^n}{\frac{r}{1 + r}} = \frac{a(1 + r)}{r} \left(1 - \frac{1}{(1 + r)^n}\right). \]

### Step 4: Find the value of \( a + a(1 + r) + \ldots + a(1 + r)^{n-1} \)

Now, we need to find the sum \( S = a + a(1 + r) + a(1 + r)^2 + \ldots + a(1 + r)^{n-1} \).

This is also a geometric series where:

- The first term \( a_1 = a \)
- The common ratio \( r = 1 + r \)
- The number of terms \( n \)

Using the geometric series formula again:

\[ S = a \frac{(1 + r)^n - 1}{(1 + r) - 1} = a \frac{(1 + r)^n - 1}{r}. \]

### Final Result

Thus, the value of \( a + a(1 + r) + a(1 + r)^2 + \ldots + a(1 + r)^{n-1} \) is:

\[ S = \frac{a((1 + r)^n - 1)}{r}. \]