Question

The $7^{th}$ term from the end in the expansion of ${(x-\frac{2}{x^2})}^{10}$ is equal to:

• A. ${}^{10}{C}_4\cdot 2^4\left(\frac{1}{x^2}\right)$
• B. ${}^{10}{C}_4\cdot 2^4$
• C. ${-}^{10}{C}_4\cdot 2^{3}x$
• D. none of the above

Answer 1

The correct option is A ${}^{10}{C}_4\cdot 2^4\left(\frac{1}{x^2}\right)$

Given

${(x-\frac{2}{x^2})}^{10}$

General term of given expansion is

$T_{r+1}={\sum }_{r=0}^{10}{}^{10}{C}_r\left(x{)}^{10-r}{\left(\frac{2}{x^2}\right)}^r\right(-1{)}^r$

We know that

The $r^{th}$ term from end is equal to $(n-r+2{)}^{th}$ term from begining

Hence

We need $7^{th}$ term from end i.e $(10-7+2{)}^{th}=5^{th}$ term from end So put $r=4$

$T_{4+1}={}^{10}{C}_4\left(x{)}^{10-4}{\left(\frac{2}{x^2}\right)}^4\right(-1{)}^4$

$T_5={}^{10}{C}_4(x{)}^6\left(\frac{2^4}{x^8}\right)$

$T_5={}^{10}{C}_4\cdot 2^4\left(\frac{1}{x^2}\right)$