Find the 7th term n the expansion of `(3x^2-1/(x^3))^(10)dot`

Answer

To find the 7th term in the expansion of \((3x^2 - \frac{1}{x^3})^{10}\), we can use the Binomial Theorem. The general term in the expansion of \((a + b)^n\) is given by:

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

In our case, \(a = 3x^2\), \(b = -\frac{1}{x^3}\), and \(n = 10\).

### Step 1: Write the general term
The general term \(T_{r+1}\) can be expressed as:

\[ T_{r+1} = \binom{10}{r} (3x^2)^{10-r} \left(-\frac{1}{x^3}\right)^r \]

### Step 2: Simplify the general term
Now, let's simplify this expression:

\[ T_{r+1} = \binom{10}{r} (3^{10-r} (x^2)^{10-r}) \left(-1\right)^r \left(\frac{1}{(x^3)^r}\right) \]

This simplifies to:

\[ T_{r+1} = \binom{10}{r} (-1)^r 3^{10-r} x^{2(10-r)} \cdot x^{-3r} \]

Combining the powers of \(x\):

\[ T_{r+1} = \binom{10}{r} (-1)^r 3^{10-r} x^{20 - 2r - 3r} = \binom{10}{r} (-1)^r 3^{10-r} x^{20 - 5r} \]

### Step 3: Find the 7th term
To find the 7th term, we need to set \(r = 6\) (since \(T_{r+1}\) corresponds to \(r+1\)):

\[ T_{7} = \binom{10}{6} (-1)^6 3^{10-6} x^{20 - 5 \cdot 6} \]

### Step 4: Calculate the values
Calculating the binomial coefficient and powers:

\[ T_{7} = \binom{10}{6} \cdot 1 \cdot 3^4 \cdot x^{20 - 30} \]

Now, calculate \(\binom{10}{6}\):

\[ \binom{10}{6} = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]

Now calculate \(3^4\):

\[ 3^4 = 81 \]

Putting it all together:

\[ T_{7} = 210 \cdot 81 \cdot x^{-10} \]

### Step 5: Final result
Thus, the 7th term in the expansion is:

\[ T_{7} = \frac{210 \cdot 81}{x^{10}} = \frac{17010}{x^{10}} \]