Question

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

Answer 1

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

Total terms = 26

${11}^{th}$ term from last = ${16}^{th}$ term from start.

$\Rightarrow n=15$

$t_{n+1}{=}^m{C}_n.a^{m-n}{b}^n$ in $(a+b{)}^m$

$\therefore t_{16}{=}^{26}{C}_{15}\left(2n{)}^{11}\right(\frac{-1}{x^2}{)}^{15}$

$={-}^{26}{C}_{15}\times 2^{11}\times \frac{x^{11}}{x^{30}}$

$={-}^{26}{C}_{15}\times 2^{11}\times x^{-19}$