JEE MAIN - Mathematics (2022 - 25th June Evening Shift - No. 18)
Explanation
Given, Binomial Expansion
$${\left( {2{x^3} + {3 \over x}} \right)^{10}}$$
General term
$${T_{r + 1}} = {}^{10}{C_r}\,.\,{(2{x^3})^{10 - r}}\,.\,{\left( {{3 \over x}} \right)^r}$$
$$ = {}^{10}{C_r}\,.\,{2^{10 - r}}\,.\,{3^r}\,.\,{x^{30 - 3r}}\,.\,{x^{ - r}}$$
$$ = {}^{10}{C_r}\,.\,{2^{10 - r}}\,.\,{3^r}\,.\,{x^{30 - 4r}}$$
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
$$ = {}^{10}{C_0}\,.\,{2^{10}}\,.\,{3^0} + {}^{10}{C_1}\,.\,{2^9}\,.\,{3^1} + {}^{10}{C_2}\,.\,{2^8}\,.\,{3^2} + {}^{10}{C_3}\,.\,{2^7}\,.\,{3^3} + {}^{10}{C_4}\,.\,{2^6}\,.\,{3^4} + {}^{10}{C_5}\,.\,{2^5}\,.\,{3^5} + {}^{10}{C_6}\,.\,{2^4}\,.\,{3^6} + {}^{10}{C_7}\,.\,{2^3}\,.\,{3^7}$$
$$ = {}^{10}{C_{10}}\,.\,{2^{10}}\,.\,{3^0} + {}^{10}{C_1}\,.\,{2^9}\,.\,{3^1}\,\, + \,\,.....\,\, + \,\,{}^{10}{C_{10}}\,.\,{2^0}\,.\,{3^{10}} - \left[ {{}^{10}{C_8}\,.\,{2^2}\,.\,{3^8} + {}^{10}{C_9}\,.\,2\,.\,{3^9} + {}^{10}{C_{10}}\,.\,{2^0}\,.\,{3^{10}}} \right]$$
$$ = {(2 + 3)^{10}} - \left[ {45\,.\,4\,.\,{3^8} + 10\,.\,2\,.\,{3^9} + 1\,.\,1\,.\,{3^{10}}} \right]$$
$$ = {5^{10}} - \left[ {60 \times {3^9} + 20\,.\,{3^9} + 3\,.\,{3^9}} \right]$$
$$ = {5^{10}} - \left( {60 + 20 + 3} \right){3^9}$$
$$ = {5^{10}} - 83\,.\,{3^9}$$
$$\therefore$$ $$\beta = 83$$
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