`sum_(i=1)^(oo)sum_(j=1)^(oo)sum_(k=1)^(oo)(1)/(a^(i+j+k))` is equal to (where `|a| gt 1`)

Options

A. `(a-1)^(-3)`
B. `(3)/(a-1)`
C. `(3)/(a^(3)-1)`
D. None of these

Explanation

To solve the problem, we need to evaluate the triple summation:

\[ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{a^{i+j+k}} \]

where \( |a| > 1 \).

### Step 1: Rewrite the summation

We can rewrite the term \(\frac{1}{a^{i+j+k}}\) as follows:

\[ \frac{1}{a^{i+j+k}} = \frac{1}{a^i} \cdot \frac{1}{a^j} \cdot \frac{1}{a^k} \]

Thus, we can express the triple summation as:

\[ \sum_{i=1}^{\infty} \frac{1}{a^i} \sum_{j=1}^{\infty} \frac{1}{a^j} \sum_{k=1}^{\infty} \frac{1}{a^k} \]

### Step 2: Evaluate each summation

Each of these summations is a geometric series. The general form of a geometric series is:

\[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r} \]

For our case, we start from \(n=1\), so we have:

\[ \sum_{i=1}^{\infty} \frac{1}{a^i} = \frac{\frac{1}{a}}{1 - \frac{1}{a}} = \frac{1/a}{(a-1)/a} = \frac{1}{a-1} \]

Similarly, we have:

\[ \sum_{j=1}^{\infty} \frac{1}{a^j} = \frac{1}{a-1} \]

\[ \sum_{k=1}^{\infty} \frac{1}{a^k} = \frac{1}{a-1} \]

### Step 3: Combine the results

Now we can combine the results of the three summations:

\[ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{a^{i+j+k}} = \left(\frac{1}{a-1}\right) \cdot \left(\frac{1}{a-1}\right) \cdot \left(\frac{1}{a-1}\right) = \frac{1}{(a-1)^3} \]

### Final Result

Thus, the value of the triple summation is:

\[ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{a^{i+j+k}} = \frac{1}{(a-1)^3} \]