JEE Main 2022 — Binomial Theorem Question with Solution

Given, Binomial Expansion

$${\left(2x^3+\frac{3}{x}\right)}^{10}$$

General term

$$T_{r+1}={}^{10}{C}_r.(2x^3{)}^{10-r}.{\left(\frac{3}{x}\right)}^r$$

$$={}^{10}{C}_r.2^{10-r}.3^r.x^{30-3r}.x^{-r}$$

$$={}^{10}{C}_r.2^{10-r}.3^r.x^{30-4r}$$

For positive even power of x, 30 $$-$$ 4r should be even and positive.

For r = 0, 30 $$-$$ 4 $$\times$$ 0 = 30 (even and positive)

For r = 1, 30 $$-$$ 4 $$\times$$ 1 = 26 (even and positive)

For r = 2, 30 $$-$$ 4 $$\times$$ 2 = 22 (even and positive)

For r = 3, 30 $$-$$ 4 $$\times$$ 3 = 18 (even and positive)

For r = 4, 30 $$-$$ 4 $$\times$$ 4 = 14 (even and positive)

For r = 5, 30 $$-$$ 4 $$\times$$ 5 = 10 (even and positive)

For r = 6, 30 $$-$$ 4 $$\times$$ 6 = 6 (even and positive)

For r = 7, 30 $$-$$ 4 $$\times$$ 7 = 2 (even and positive)

For r = 8, 30 $$-$$ 4 $$\times$$ 8 = $$-$$2 (even but not positive)

So, for r = 1, 2, 3, 4, 5, 6 and 7 we can get positive even power of x.

$$\therefore$$ Sum of coefficient for positive even power of x

$$={}^{10}{C}_0.2^{10}.3^0+{}^{10}{C}_1.2^9.3^1+{}^{10}{C}_2.2^8.3^2+{}^{10}{C}_3.2^7.3^3+{}^{10}{C}_4.2^6.3^4+{}^{10}{C}_5.2^5.3^5+{}^{10}{C}_6.2^4.3^6+{}^{10}{C}_7.2^3.3^7$$

$$={}^{10}{C}_{10}.2^{10}.3^0+{}^{10}{C}_1.2^9.3^1+.....+{}^{10}{C}_{10}.2^0.3^{10}-\left[{}^{10}{C}_8.2^2.3^8+{}^{10}{C}_9.2.3^9+{}^{10}{C}_{10}.2^0.3^{10}\right]$$

$$=(2+3{)}^{10}-\left[45.4.3^8+10.2.3^9+1.1.3^{10}\right]$$

$$=5^{10}-\left[60\times 3^9+20.3^9+3.3^9\right]$$

$$=5^{10}-\left(60+20+3\right)3^9$$

$$=5^{10}-83.3^9$$

$$\therefore$$ $$\beta =83$$