find the sum (x + 1/x)^3 + (x^2 + 1/x^2)^3 +.....................+(x^n + 1/x^n)^3

To find the sum of the expression \((x + \frac{1}{x})^3 + (x^2 + \frac{1}{x^2})^3 + \ldots + (x^n + \frac{1}{x^n})^3\), we can break it down step by step. This involves recognizing patterns in the powers of \(x\) and their reciprocals, and how they relate to each other. Let's delve into the details.

Understanding the Components

First, we need to analyze the term \((x^k + \frac{1}{x^k})^3\) for any integer \(k\). We can use the binomial expansion to simplify this expression:

Binomial Expansion

The binomial expansion of \((a + b)^3\) is given by:

• \(a^3 + 3a^2b + 3ab^2 + b^3\)

In our case, let \(a = x^k\) and \(b = \frac{1}{x^k}\). Thus, we have:

\((x^k + \frac{1}{x^k})^3 = (x^k)^3 + 3(x^k)^2(\frac{1}{x^k}) + 3(x^k)(\frac{1}{x^k})^2 + (\frac{1}{x^k})^3\)

This simplifies to:

\(x^{3k} + 3x^{k} + 3\frac{1}{x^{k}} + \frac{1}{x^{3k}}\)

Summing the Series

Now, we can express the entire sum as follows:

\(\sum_{k=1}^{n} (x^k + \frac{1}{x^k})^3 = \sum_{k=1}^{n} \left( x^{3k} + 3x^{k} + 3\frac{1}{x^{k}} + \frac{1}{x^{3k}} \right)\)

This can be separated into four distinct sums:

• \(\sum_{k=1}^{n} x^{3k}\)
• \(3\sum_{k=1}^{n} x^{k}\)
• \(3\sum_{k=1}^{n} \frac{1}{x^{k}}\)
• \(\sum_{k=1}^{n} \frac{1}{x^{3k}}\)

Calculating Each Sum

Each of these sums can be calculated using the formula for the sum of a geometric series:

\(S_n = a \frac{1 - r^n}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio.

First Sum: \( \sum_{k=1}^{n} x^{3k} \)

Here, \(a = x^3\) and \(r = x^3\):

\(\sum_{k=1}^{n} x^{3k} = x^3 \frac{1 - (x^3)^n}{1 - x^3} = \frac{x^3(1 - x^{3n})}{1 - x^3}\)

Second Sum: \( 3\sum_{k=1}^{n} x^{k} \)

For this sum, \(a = x\) and \(r = x\):

\(3\sum_{k=1}^{n} x^{k} = 3x \frac{1 - x^n}{1 - x}\)

Third Sum: \( 3\sum_{k=1}^{n} \frac{1}{x^{k}} \)

In this case, \(a = \frac{1}{x}\) and \(r = \frac{1}{x}\):

\(3\sum_{k=1}^{n} \frac{1}{x^{k}} = 3 \frac{\frac{1}{x}(1 - \frac{1}{x^n})}{1 - \frac{1}{x}} = \frac{3(1 - \frac{1}{x^n})}{x - 1}\)

Fourth Sum: \( \sum_{k=1}^{n} \frac{1}{x^{3k}} \)

Here, \(a = \frac{1}{x^3}\) and \(r = \frac{1}{x^3}\):

\(\sum_{k=1}^{n} \frac{1}{x^{3k}} = \frac{\frac{1}{x^3}(1 - \frac{1}{x^{3n}})}{1 - \frac{1}{x^3}} = \frac{\frac{1}{x^3}(1 - \frac{1}{x^{3n}})}{\frac{x^3 - 1}{x^3}} = \frac{1 - \frac{1}{x^{3n}}}{x^3 - 1}\)

Final Expression

Combining all these results gives us the complete sum:

\(S = \frac{x^3(1 - x^{3n})}{1 - x^3} + 3x \frac{1 - x^n}{1 - x} + \frac{3(1 - \frac{1}{x^n})}{x - 1} + \frac{1 - \frac{1}{x^{3n}}}{x^3 - 1}\)

This expression encapsulates the sum of the series you were interested in. Each component reflects the contributions from the various powers of \(x\) and their reciprocals, showcasing the beauty of algebraic manipulation and the power of geometric series.