JEE Main 2022 — Binomial Theorem Question with Solution

Here property used is

$${}^n{C}_r+{}^n{C}_{r+1}={}^{n+1}{C}_{r+1}$$

Given, $${}^{40}{C}_0+{}^{41}{C}_1+{}^{42}{C}_2+....+{}^{60}{C}_{20}=\frac{m}{n}{}^{60}{C}_{20}$$

As $${}^{40}{C}_0={}^{41}{C}_0=1$$

So, we replace $${}^{40}{C}_0$$ with $${}^{41}{C}_0$$.

$$\Rightarrow {}^{41}{C}_0+{}^{41}{C}_1+{}^{42}{C}_2+.....+{}^{60}{C}_{20}=\frac{m}{n}.{}^{60}{C}_{20}$$

$$\Rightarrow {}^{42}{C}_1+{}^{42}{C}_2+.....+{}^{60}{C}_{20}=\frac{m}{n}.{}^{60}{C}_{20}$$

$$\Rightarrow {}^{43}{C}_2+{}^{43}{C}_3+.....+{}^{60}{C}_{20}=\frac{m}{n}.{}^{60}{C}_{20}$$

$$\Rightarrow {}^{44}{C}_3+{}^{44}{C}_4+.....+{}^{60}{C}_{20}=\frac{m}{n}.{}^{60}{C}_{20}$$

$$\Rightarrow {}^{45}{C}_4+{}^{45}{C}_5+.....+{}^{60}{C}_{20}=\frac{m}{n}.{}^{60}{C}_{20}$$

$$⋮$$

$$\Rightarrow {}^{60}{C}_{19}+{}^{60}{C}_{20}=\frac{m}{n}.{}^{60}{C}_{20}$$

$$\Rightarrow {}^{61}{C}_{20}=\frac{m}{n}.{}^{60}{C}_{20}$$

$$\Rightarrow \frac{61!}{20!41!}=\frac{m}{n}.\frac{60!}{20!40!}$$

$$\Rightarrow \frac{61}{41}=\frac{m}{n}$$

$$\therefore$$ m = 61 and n = 41

$$\therefore$$ m + n = 61 + 41 = 102