If `S_p` denote sum of the series `1+r^p+r^(2p)+ . . . oo and s_p` denote the sum `1-r^p+r^(2p)-r^(3p) . . . oo , abs r lt 1` then `S_p+s_p` equals

Options

A. `2S_(2p)`
B. 0
C. `1/2S_(2p)`
D. `-1/2S_(2p)`

Answer

To solve the problem, we need to find the sums \( S_p \) and \( s_p \) and then compute \( S_p + s_p \).

### Step 1: Calculate \( S_p \)

The series \( S_p \) is given by:

\[ S_p = 1 + r^p + r^{2p} + r^{3p} + \ldots \]

This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = r^p \). The formula for the sum of an infinite geometric series is:

\[ S = \frac{a}{1 - r} \]

Applying this formula:

\[ S_p = \frac{1}{1 - r^p} \quad \text{(for } |r| < 1\text{)} \]

### Step 2: Calculate \( s_p \)

The series \( s_p \) is given by:

\[ s_p = 1 - r^p + r^{2p} - r^{3p} + \ldots \]

This is also a geometric series where the first term \( a = 1 \) and the common ratio \( r = -r^p \). Using the same formula for the sum of an infinite geometric series:

\[ s_p = \frac{1}{1 - (-r^p)} = \frac{1}{1 + r^p} \quad \text{(for } |r| < 1\text{)} \]

### Step 3: Calculate \( S_p + s_p \)

Now, we need to find \( S_p + s_p \):

\[ S_p + s_p = \frac{1}{1 - r^p} + \frac{1}{1 + r^p} \]

To combine these fractions, we find a common denominator:

\[ S_p + s_p = \frac{(1 + r^p) + (1 - r^p)}{(1 - r^p)(1 + r^p)} = \frac{2}{1 - (r^p)^2} \]

Using the identity \( 1 - (r^p)^2 = 1 - r^{2p} \):

\[ S_p + s_p = \frac{2}{1 - r^{2p}} \]

### Final Answer

Thus, the final result is:

\[ S_p + s_p = \frac{2}{1 - r^{2p}} \]

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