Find the 11th term from the beginning and the 11th term from the end in the expansion of

Question:

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

Solution:

Given:

$\left(2 x-\frac{1}{x^{2}}\right)^{25}$

Clearly, the given expression contains 26 terms.

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

Thus, we have:

$T_{16}=T_{15+1}={ }^{25} C_{15}(2 x)^{25-15}\left(\frac{-1}{x^{2}}\right)^{15}$

$={ }^{25} C_{15}\left(2^{10}\right)\left(x^{10}\right)\left(\frac{-1}{x^{30}}\right)=-{ }^{25} C_{15}\left(\frac{2^{10}}{x^{20}}\right)$

Now, we will find the 11th term from the beginning.

$T_{11}=T_{10+1}$

$={ }^{25} C_{10}(2 x)^{25-10}\left(\frac{-1}{x^{2}}\right)^{10}$

$={ }^{25} C_{10}\left(2^{15}\right)\left(x^{15}\right)\left(\frac{1}{x^{20}}\right)$

$={ }^{25} C_{10}\left(\frac{2^{15}}{x^{5}}\right)$