General term of `(1 -x^2)^12` is ……….

Answer

To find the general term of the expression \((1 - x^2)^{12}\), 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 our case, we have:
- \(a = 1\)
- \(b = -x^2\)
- \(n = 12\)

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:

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

### Step 2: Substitute the Values
Substituting the values of \(a\), \(b\), and \(n\) into the formula, we get:

\[ T_{r+1} = \binom{12}{r} (1)^{12-r} (-x^2)^r \]

### Step 3: Simplify the Expression
Since \((1)^{12-r} = 1\), we can simplify the expression further:

\[ T_{r+1} = \binom{12}{r} (-x^2)^r \]

### Step 4: Expand the Power
Now, we can expand \((-x^2)^r\):

\[ T_{r+1} = \binom{12}{r} (-1)^r (x^2)^r \]

### Step 5: Final Form of the General Term
Thus, the general term can be expressed as:

\[ T_{r+1} = \binom{12}{r} (-1)^r x^{2r} \]

### Conclusion
The general term of the expansion of \((1 - x^2)^{12}\) is:

\[ T_{r+1} = \binom{12}{r} (-1)^r x^{2r} \]

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