Find the specified term of the expression in each of the following binomials:
(ii) Sixth term of `(2x - (1)/( x^2) )^7`.

Answer

To find the sixth term of the expression \((2x - \frac{1}{x^2})^7\), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In this case, we have \(a = 2x\), \(b = -\frac{1}{x^2}\), and \(n = 7\). The \(r+1\)th term (or the \((r+1)\)th term) in the expansion is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

To find the sixth term, we need to set \(r = 5\) (since the sixth term corresponds to \(r = 5\)).

### Step 1: Identify \(n\), \(a\), \(b\), and \(r\)
- \(n = 7\)
- \(a = 2x\)
- \(b = -\frac{1}{x^2}\)
- \(r = 5\)

### Step 2: Write the formula for the sixth term
Using the formula for the term, we can write:

\[ T_6 = T_{r+1} = T_{5+1} = \binom{7}{5} (2x)^{7-5} \left(-\frac{1}{x^2}\right)^5 \]

### Step 3: Calculate \(\binom{7}{5}\)
Using the combination formula:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

We find:

\[ \binom{7}{5} = \frac{7!}{5! \cdot 2!} = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 4: Substitute values into the term formula
Now substituting back into the term formula:

\[ T_6 = 21 (2x)^{2} \left(-\frac{1}{x^2}\right)^{5} \]

### Step 5: Simplify the expression
Calculating \( (2x)^{2} \):

\[ (2x)^{2} = 4x^{2} \]

Calculating \(\left(-\frac{1}{x^2}\right)^{5}\):

\[ \left(-\frac{1}{x^2}\right)^{5} = -\frac{1}{x^{10}} \]

Now substituting these values back into the term:

\[ T_6 = 21 \cdot 4x^{2} \cdot \left(-\frac{1}{x^{10}}\right) \]

### Step 6: Combine the terms
\[ T_6 = 21 \cdot 4 \cdot -\frac{x^{2}}{x^{10}} = -84 \cdot \frac{1}{x^{8}} = -\frac{84}{x^{8}} \]

### Final Answer
Thus, the sixth term of the expression \((2x - \frac{1}{x^2})^7\) is:

\[ T_6 = -\frac{84}{x^{8}} \]