If the coefficient of x 10 in the binomial expansion of ( x 5^1 4 + 5 x^1 3 )^60 is 5^k\,.\,l, where l, k N and l is co-prime to 5, then k is equal to _____________.

Given Binomial Expansion = ( x 5^1 4 + x 5^1 3 )^60 General term T_r + 1 = ^60C_r\,.\,( x^1/2 5^1/4 )^60 - r\,.\,( 5^1/2 x^1/3 )^r = ^60C_r\,.\,5^( r 4 - 15 + r 2 )\,.\,x^( 30 - r 2 - r 3 ) = ^60C_r\,.\,5^( 3r - 60 4 )\,.\,x^( 180 - 5r 6 ) For x 10 term, 180 - 5r 6 = 10 5r =

If the coefficient of x10 in the binomial expansion of x 51 : null (Mathematics)

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 \quad ...... (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

Explanation

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