Finding \(a^5 + \frac{1}{a^5}\) Value from \(a + \frac{1}{a}\)

The problem asks us to find the value of \(a^5 + \frac{1}{a^5}\) given that \(a + \frac{1}{a} = 7\). To solve this, we can find the values of lower power expressions like \(a^2 + \frac{1}{a^2}\) and \(a^3 + \frac{1}{a^3}\) first, and then combine them appropriately.

Step 1: Calculate \(a^2 + \frac{1}{a^2}\)

We start by squaring the given expression:

\(\left(a + \frac{1}{a}\right)^2 = a^2 + 2 \cdot a \cdot \frac{1}{a} + \left(\frac{1}{a}\right)^2\)

\(\left(a + \frac{1}{a}\right)^2 = a^2 + 2 + \frac{1}{a^2}\)

Rearranging the terms to find \(a^2 + \frac{1}{a^2}\):

\(a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2\)

Substitute the given value \(a + \frac{1}{a} = 7\):

\(a^2 + \frac{1}{a^2} = (7)^2 - 2\)

\(a^2 + \frac{1}{a^2} = 49 - 2\)

\(a^2 + \frac{1}{a^2} = 47\)

Step 2: Calculate \(a^3 + \frac{1}{a^3}\)

Next, we cube the given expression:

\(\left(a + \frac{1}{a}\right)^3 = a^3 + 3 \cdot a \cdot \frac{1}{a} \left(a + \frac{1}{a}\right) + \left(\frac{1}{a}\right)^3\)

\(\left(a + \frac{1}{a}\right)^3 = a^3 + 3\left(a + \frac{1}{a}\right) + \frac{1}{a^3}\)

Rearranging the terms to find \(a^3 + \frac{1}{a^3}\):

\(a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right)\)

Substitute the given value \(a + \frac{1}{a} = 7\):

\(a^3 + \frac{1}{a^3} = (7)^3 - 3(7)\)

\(a^3 + \frac{1}{a^3} = 343 - 21\)

\(a^3 + \frac{1}{a^3} = 322\)

Step 3: Calculate \(a^5 + \frac{1}{a^5}\)

We can express \(a^5 + \frac{1}{a^5}\) using the values of \(a^2 + \frac{1}{a^2}\) and \(a^3 + \frac{1}{a^3}\). Consider the product of these two expressions:

\(\left(a^2 + \frac{1}{a^2}\right)\left(a^3 + \frac{1}{a^3}\right) = a^2 \cdot a^3 + a^2 \cdot \frac{1}{a^3} + \frac{1}{a^2} \cdot a^3 + \frac{1}{a^2} \cdot \frac{1}{a^3}\)

Simplify the terms:

\(= a^{2+3} + a^{2-3} + a^{3-2} + a^{-2-3}\)

\(= a^5 + a^{-1} + a^1 + a^{-5}\)

\(= a^5 + \frac{1}{a} + a + \frac{1}{a^5}\)

Rearrange the terms:

\(= \left(a^5 + \frac{1}{a^5}\right) + \left(a + \frac{1}{a}\right)\)

So, we have the relationship:

\(\left(a^2 + \frac{1}{a^2}\right)\left(a^3 + \frac{1}{a^3}\right) = \left(a^5 + \frac{1}{a^5}\right) + \left(a + \frac{1}{a}\right)\)

To find \(a^5 + \frac{1}{a^5}\), we rearrange this equation:

\(a^5 + \frac{1}{a^5} = \left(a^2 + \frac{1}{a^2}\right)\left(a^3 + \frac{1}{a^3}\right) - \left(a + \frac{1}{a}\right)\)

Substitute the values we calculated in Step 1 and Step 2, and the given value:

\(a^5 + \frac{1}{a^5} = (47)(322) - 7\)

First, calculate the product \(47 \times 322\):

\(47 \times 322 = 15134\)

Now, complete the calculation for \(a^5 + \frac{1}{a^5}\):

\(a^5 + \frac{1}{a^5} = 15134 - 7\)

\(a^5 + \frac{1}{a^5} = 15127\)

Thus, the value of \(a^5 + \frac{1}{a^5}\) is 15127.

Revision Table: Key Values

Additional Information: Generalizing for Higher Powers

We can find the values of \(a^n + \frac{1}{a^n}\) for higher integer values of \(n\) if we know \(a + \frac{1}{a}\). Let \(x_n = a^n + \frac{1}{a^n}\). Then we have the relation:

• \(x_1 = a + \frac{1}{a}\) (given)
• \(x_2 = \left(a + \frac{1}{a}\right)^2 - 2 = x_1^2 - 2\)
• \(x_3 = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right) = x_1^3 - 3x_1\)

For \(n \ge 2\), we can use the recursive relation:

\(x_n = x_{n-1} \cdot x_1 - x_{n-2}\)

Let's verify this for \(n=3\):

\(x_3 = x_2 \cdot x_1 - x_1\)

\(a^3 + \frac{1}{a^3} = \left(a^2 + \frac{1}{a^2}\right)\left(a + \frac{1}{a}\right) - \left(a + \frac{1}{a}\right)\)

\(\left(a^2 + \frac{1}{a^2}\right)\left(a + \frac{1}{a}\right) = a^3 + a + \frac{1}{a} + \frac{1}{a^3} = \left(a^3 + \frac{1}{a^3}\right) + \left(a + \frac{1}{a}\right)\)

So, \(\left(a^3 + \frac{1}{a^3}\right) = \left(a^2 + \frac{1}{a^2}\right)\left(a + \frac{1}{a}\right) - \left(a + \frac{1}{a}\right)\). This confirms the formula for \(n=3\).

Using this general formula for \(n=5\):

\(x_5 = x_4 \cdot x_1 - x_3\)

We could calculate \(x_4\) first: \(x_4 = x_3 \cdot x_1 - x_2\). With \(x_1=7\), \(x_2=47\), \(x_3=322\): \(x_4 = (322)(7) - 47 = 2254 - 47 = 2207\)

Then, \(x_5 = x_4 \cdot x_1 - x_3\):

\(x_5 = (2207)(7) - 322 = 15449 - 322 = 15127\)

This confirms the result obtained using the \(x_2 \cdot x_3\) method. Both methods are valid for finding \(a^5 + \frac{1}{a^5}\).