JEE Main 2022 — Binomial Theorem Question with Solution

Given,

$$9^n-8n-1=64\alpha$$

$$\Rightarrow \alpha =\frac{{(1+8)}^n-8n-1}{64}$$

$$=\frac{\left({}^n{C}_0.1+{}^n{C}_1.8^1+{}^n{C}_2.8^2+.....+{}^n{C}_n.8^n\right)-8n-1}{8^2}$$

$$=\frac{1+8n+{}^n{C}_2.8^2+....+{}^n{C}_n.8^n-8n-1}{8^2}$$

$$=\frac{{}^n{C}_2.8^2+{}^n{C}_3.8^3+.....+{}^n{C}_n.8^n}{8^2}$$

$$={}^n{C}_2+{}^n{C}_3.8+{}^n{C}_4.8^2+.....{}^n{C}_n.8^{n-2}$$

Also given,

$$6^n-5n-1=25\beta$$

$$\Rightarrow \beta =\frac{{(1+5)}^n-5n-1}{25}$$

$$=\frac{{}^n{C}_0.1+{}^n{C}_1.5+{}^n{C}_2.5^2+.....+{}^n{C}_n.5^n-5n-1}{5^2}$$

$$=\frac{1+5n+{}^n{C}_2.5^2+{}^n{C}_3.5^3+....+{}^n{C}_n.5^2-5n-1}{5^2}$$

$$=\frac{{}^n{C}_2.5^2+{}^n{C}_3.5^3+{}^n{C}_4.5^4+....+{}^n{C}_n.5^n}{5^2}$$

$$={}^n{C}_2+{}^n{C}_3.5+{}^n{C}_4.5^2+.....+{}^n{C}_n.5^{n-2}$$

$$\therefore$$ $$\alpha -\beta$$

$$=\left({}^n{C}_2+{}^n{C}_3.8+{}^n{C}_4.8^2+....+{}^n{C}_n.8^{n-2}\right)-\left({}^n{C}_2+{}^n{C}_3.5+{}^n{C}_4.5^2+....+{}^n{C}_n.5^{n-2}\right)$$

$$={}^n{C}_3.(8-5)+{}^n{C}_4.(8^2-5^2)+....+{}^n{C}_n(8^{n-2}-5^{n-2})$$