In `(3x^(2)-(1)/(x))^(16)` which term contains `x^(12)` .

Answer

To find the term that contains \( x^{12} \) in the expansion of \( (3x^2 - \frac{1}{x})^{16} \), 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, \( a = 3x^2 \), \( b = -\frac{1}{x} \), and \( n = 16 \). Therefore, the general term becomes:

\[ T_{r+1} = \binom{16}{r} (3x^2)^{16-r} \left(-\frac{1}{x}\right)^r \]

Now, we simplify this expression:

\[ T_{r+1} = \binom{16}{r} (3^{16-r} (x^2)^{16-r}) \left(-1^r \frac{1}{x^r}\right) \]

This can be rewritten as:

\[ T_{r+1} = \binom{16}{r} (-1)^r 3^{16-r} x^{2(16-r)} x^{-r} \]

Combining the powers of \( x \):

\[ T_{r+1} = \binom{16}{r} (-1)^r 3^{16-r} x^{32 - 2r - r} = \binom{16}{r} (-1)^r 3^{16-r} x^{32 - 3r} \]

We want to find the term where the exponent of \( x \) is \( 12 \):

\[ 32 - 3r = 12 \]

Now, we solve for \( r \):

\[ 32 - 12 = 3r \\ 20 = 3r \\ r = \frac{20}{3} \]

Since \( r \) must be a whole number (as it represents the term number), we see that \( r = \frac{20}{3} \) is not an integer. Therefore, there is no term in the expansion of \( (3x^2 - \frac{1}{x})^{16} \) that contains \( x^{12} \).