QJEEM22Ma8d7c: JEE Main 2022 Mathematics - Class 11 JEEM22Ma8d7c Solution: If the coefficient of x 10 in the binomial expansion of is ,…

Given Binomial Expansion

$= {\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt x } \over {{5^{{1 \over 3}}}}}} \right)^{60}}$

$\therefore$ General term

${T_{r + 1}} = {}^{60}{C_r}\,.\,{\left( {{{{x^{1/2}}} \over {{5^{1/4}}}}} \right)^{60 - r}}\,.\,{\left( {{{{5^{1/2}}} \over {{x^{1/3}}}}} \right)^r}$

$= {}^{60}{C_r}\,.\,{5^{\left( {{r \over 4} - 15 + {r \over 2}} \right)}}\,.\,{x^{\left( {30 - {r \over 2} - {r \over 3}} \right)}}$

$= {}^{60}{C_r}\,.\,{5^{\left( {{{3r - 60} \over 4}} \right)}}\,.\,{x^{\left( {{{180 - 5r} \over 6}} \right)}}$

For x¹⁰ term,

${{180 - 5r} \over 6} = 10$

$\Rightarrow 5r = 120$

$\Rightarrow r = 24$

$\therefore$ Coefficient of ${x^{10}} = {}^{60}{C_{24}}\,.\,{5^{\left( {{{3 \times 24 - 60} \over 4}} \right)}}$

$= {}^{60}{C_{24}}\,.\,{5^3}$

$= {{60!} \over {24!\,\,36!}}\,.\,{5^3}$

It is given that,

${{60!} \over {24!\,\,36!}}\,.\,{5^3} = {5^k}\,.\,l$ ...... (1)

Also given that, l is coprime to 5 means l can't be multiple of 5. So we have to find all the factors of 5 in 60!, 24! and 36!

[Note : Formula for exponent or degree of prime number in n!.

Exponent of p in $n! = \left\lceil {{n \over p}} \right\rceil + \left\lceil {{n \over {{p^2}}}} \right\rceil + \left\lceil {{n \over {{p^3}}}} \right\rceil +$ ..... until 0 comes

here p is a prime number. ]

$\therefore$ Exponent of 5 in 60!

$= \left\lceil {{{60} \over 5}} \right\rceil + \left\lceil {{{60} \over {{5^2}}}} \right\rceil + \left\lceil {{{60} \over {{5^3}}}} \right\rceil +$ .....

$= 12 + 2 + 0 +$ .....

$= 14$

Exponent of 5 in 24!

$= \left\lceil {{{24} \over 5}} \right\rceil + \left\lceil {{{24} \over {{5^2}}}} \right\rceil + \left\lceil {{{24} \over {{5^3}}}} \right\rceil +$ ......

$= 4 + 0 + 0$ ......

$= 4$

Exponent of 5 in 36!

$= \left\lceil {{{36} \over 5}} \right\rceil + \left\lceil {{{36} \over {{5^2}}}} \right\rceil + \left\lceil {{{36} \over {{5^3}}}} \right\rceil +$ .......

$= 7 + 1 + 0$ ......

$= 8$

$\therefore$ From equation (1), exponent of 5 overall

${{{5^{14}}} \over {{5^4}\,.\,{5^8}}}\,.\,{5^3} = {5^k}$

$\Rightarrow {5^5} = {5^k}$

$\Rightarrow k = 5$