Question

The ratio of 11th terms from the beginning and ${11}^{th}$ terms from the end in the expansion of ${(2x-\frac{1}{x^2})}^{25}$ is

• A. $1$
• B. $-2^5{x}^{15}$
• C. $2^{15}$
• D. $-\frac{2^{15}}{x^5}$

Answer 1

Answer: (B)

$-2^5{x}^{15}$

The (r+1)th term in the binomial expansion of (x+a)${}^n$ is $T_{r+1}{=}^n{C}_r{x}^{n-r}{a}^r$
$\therefore$11th term from the beginning $=T_{11}=T_{10+1}$
${=}^{25}{C}_{10}(2x{)}^{25-10}{(-\frac{1}{x^2})}^{10}{=}^{25}{C}_{10}{2}^{15}\cdot \frac{1}{x^5}$
And 11 th term from the end $=$(26 - 10)th $=$16th term from the beginning
${=}^{25}{C}_{15}\left(2x\right){(-\frac{1}{x^2})}^{15}{=}^{25}{C}_{15}{2}^{10}\frac{1}{x^{20}}$
Hence, the desired ration $=\frac{{}^{25}{C}_{10}{2}^{15}\frac{1}{x^5}}{{}^{25}{C}_{15}{2}^{10}\frac{1}{x^{20}}}=-2^5{x}^{15}$