JEE Main 2022 — Binomial Theorem Question with Solution

General term of Binomial expansion $${\left(\frac{5}{2}{x}^3-\frac{1}{5x^2}\right)}^{11}$$ is

$$T_{r+1}={}^{11}{C}_r.{\left(\frac{5}{2}{x}^3\right)}^{11-r}.{\left(-\frac{1}{5x^2}\right)}^r$$

$$={}^{11}{C}_r.{\left(\frac{5}{2}\right)}^{11-r}.{\left(-\frac{1}{5}\right)}^r.x^{33-5r}$$

In the term,

$$\left(1-x^2+3x^3\right){\left(\frac{5}{2}{x}^3-\frac{1}{5x^2}\right)}^{11}$$

Term independent of x is when

(1) $$33-5r=0$$

$$\Rightarrow r=\frac{33}{5}\notin$$ integer

(2) $$33-5r=-2$$

$$\Rightarrow 5r=35$$

$$\Rightarrow r=7\in$$ integer

(3) $$33-5r=-3$$

$$\Rightarrow 5r=36$$

$$\Rightarrow r=\frac{36}{5}\notin$$ integer

$$\therefore$$ Only for r = 7 independent of x term possible.

$$\therefore$$ Independent of x term

$$=-\left({}^{11}{C}_7{\left(\frac{5}{2}\right)}^4.{\left(-\frac{1}{5}\right)}^7\right)$$

$$=-\left(\frac{11.10.9.8}{4.3.2.1}.\frac{5^4}{2^4}.-\frac{1}{5^7}\right)$$

$$=\frac{11.10.3}{2^4.5^3}$$

$$=\frac{11.3}{2^3.5^2}$$

$$=\frac{33}{200}$$