Question

In the expansion of ${(x^3-\frac{1}{x^2})}^{15}$, the constant terms is

• A. ${}^{15}{C}_6$
• B. $-{}^{15}{C}_6$
• C. ${}^{15}{C}_4$
• D. $-{}^{15}{C}_4$

Answer 1

The correct option is B $-{}^{15}{C}_6$

We have,

${\left(x^2-\frac{1}{x^2}\right)}^{15}{T}_{r+1}{=}^{15}{C}_r{\left(-1\right)}^r{x}^{15-r}{=}^{15}{C}_r{\left(-1\right)}^r{x}^{45-5r}$

$Forconstantterm45-5t=0r=9$

Therefore,

$={-}^{15}{C}_9={-}^{15}{C}_6$

Hence, this is the answer.