"The number of positive integers k such that the constant term in the binomial expansion of ${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$, x $\ne$ 0 is 28 . l, where l is an odd integer, is ______________."

Question

Accepted Answer

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Given Binomial expression is

${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$

General term,

$${T_{r + 1}} = {}^{12}{C_r}{(2{x^3})^r}\,.\,{\left( {{3 \over {{x^k}}}} \right)^{12 - r}}$$

$$ = \left( {{}^{12}{C_r}\,.\,{2^r}\,.\,{3^{12 - r}}} \right)\,.\,{x^{3r - 12k + kr}}$$

For constant term,

$3r - 12k + kr = 0$

$\Rightarrow k(12 - r) = 3r$

$\Rightarrow k = {{3r} \over {12 - r}}$

For r = 1, $k = {3 \over {11}}$ (not integer)

For r = 2, $k = {6 \over {10}}$ (not integer)

For r = 3, $k = {9 \over {9}}=1$ (integer)

For r = 6, $k = {18 \over {6}}=3$ (integer)

For r = 8, $k = {24 \over {4}}=6$ (integer)

For r = 9, $k = {27 \over {3}}=9$ (integer)

For r = 10, $k = {30 \over {2}}=15$ (integer)

For r = 11, $k = {33 \over {1}}=33$ (integer)

So, for r = 3, 6, 8, 9, 10 and 11 k is positive integer.

When k = 1 then r = 3 and constant term is

$= {}^{12}{C_3}\,.\,{2^3}\,.\,{3^9}$

$= {{12\,.\,11\,.\,10} \over {3\,.\,2\,.\,1}}\,.\,{2^3}\,.\,{3^9}$

$= 2\,.\,11\,.\,2\,.\,5\,.\,{2^3}\,.\,{3^9}$

$= 11\,.\,5\,.\,{2^5}\,.\,{3^9}$

$= {2^5}\,.\,(55\,.\,{3^9})$

$= {2^5}(l)$

$\ne {2^8}\,.\,l$

When x = 3 then r = 6 and constant term

$= {}^{12}{C_6}\,.\,{2^6}\,.\,{3^6}$

$$ = {{12\,.\,11\,.\,10\,.\,9\,.\,8\,.\,7} \over {6\,.\,5\,.\,4\,.\,3\,.\,2\,.\,1}}\,.\,{2^6}\,.\,{3^6}$$

$= {2^8}\,.\,231\,.\,{3^6}$

$= {2^8}(l)$

When k = 6 then r = 8 and constant term

$= {}^{12}{C_8}\,.\,{2^8}\,.\,{3^4}$

$= {{12\,.\,11\,.\,10\,.\,9} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{2^8}\,.\,{3^4}$

$= {2^8}\,.\,55\,.\,{3^6}$

$= {2^8}(l)$

When x = 9 then r = 9 and constant term

$= {}^{12}{C_9}\,.\,{2^9}\,.\,{3^3}$

$= {{12\,.\,11\,.\,10} \over {3\,.\,2\,.\,1}}\,.\,{2^9}\,.\,{3^3}$

$= {2^{11}}\,.\,55\,.\,{3^3}$

Here power of 2 is 11 which is greater than 8. So, k = 9 is not possible.

Similarly for k = 15 and k = 33, ${2^8}\,.\,l$ form is not possible.

$\therefore$ k = 3 and k = 6 is accepted.

$\therefore$ For 2 positive integer value of k, ${2^8}\,.\,l$ form of constant term possible.

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The number of positive integers k such that the constant term in the binomial expansion of ${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$, x $\ne$ 0 is 2⁸ . l, where l is an odd integer, is ______________.

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The number of positive integers k such that the constant term in the binomial expansion of ${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$, x $\ne$ 0 is 28 . l, where l is an odd integer, is ______________.

Solution

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The number of positive integers k such that the constant term in the binomial expansion of ${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$, x $\ne$ 0 is 2⁸ . l, where l is an odd integer, is ______________.