Question

Find the 11th term from the beginning and the 11th term from the end in the expansion of ${\left(2x-\frac{1}{x^2}\right)}^{25}$.

Answer 1

Given:
${\left(2x-\frac{1}{x^2}\right)}^{25}$
Clearly, the given expression contains 26 terms.

So, the 11^th term from the end is the (26 − 11 + 1)^th term from the beginning. In other words, the 11^th term from the end is the 16^th term from the beginning.

Thus, we have:
$$\begin{gathered} T_{16}=T_{15+1}={}^{25}C_{15}(2x{)}^{25-15}{\left(\frac{-1}{x^2}\right)}^{15} \\ ={}^{25}C_{15}\left(2^{10}\right)\left(x^{10}\right)\left(\frac{-1}{x^{30}}\right)=-{}^{25}C_{15}\left(\frac{2^{10}}{x^{20}}\right) \end{gathered}$$
Now, we will find the 11^th term from the beginning.

$$\begin{gathered} T_{11}=T_{10+1} \\ ={}^{25}C_{10}(2x{)}^{25-10}{\left(\frac{-1}{x^2}\right)}^{10} \\ ={}^{25}C_{10}\left(2^{15}\right)\left(x^{15}\right)\left(\frac{1}{x^{20}}\right) \\ ={}^{25}C_{10}\left(\frac{2^{15}}{x^5}\right) \end{gathered}$$