Question

Find the 5th term from the end in the expansion of ${\left(3x-\frac{1}{x^2}\right)}^{10}$

Answer 1

Given:
${\left(3x-\frac{1}{x^2}\right)}^{10}$
Clearly, the expression has 6 terms.
The 5th term from the end is the (11 − 5 + 1)th, i.e., 7th, term from the beginning.
Thus, we have:

$$\begin{gathered} T_7=T_{6+1} \\ ={}^{10}C_6(3x{)}^{10-6}{\left(\frac{-1}{x^2}\right)}^6 \\ ={}^{10}C_6\left(3^4\right)\left(x^4\right)\left(\frac{1}{x^{12}}\right) \\ =\frac{10\times 9\times 8\times 7\times 81}{4\times 3\times 2\times 1\times x^8}=\frac{17010}{x^8} \end{gathered}$$