Let $(x^{3} + x^{5}) {(2 x^{6} + 3 x^{4} - 1)}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + ........ + a_{125} x^{125}$ , then which of the following is/are TRUE?

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$a_{1} + a_{2} + a_{3} = 1$

$a_{125} = 2^{20}$

$a_{0} - \frac{a_{2}}{3} + \frac{a_{4}}{5} - \frac{a_{6}}{7} + ...... + \frac{a_{124}}{125} = 0$

$a_{1} + a_{3} + a_{5} + ...... + a_{123} = 2^{41} - 2^{20}$

answer is A, B, C, D.

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Detailed Solution

$(x^{3} + x^{5}) {(2 x^{6} + 3 x^{4} - 1)}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{125} x^{125}$

In the expansion of x is obviously zero.
Put x = 1, and x = -1 and subtract we get
$2 (a_{1} + a_{3} + ...... + a_{125}) = {2.4}^{20} - (- 2) {.4}^{20} \Rightarrow a_{1} + a_{3} + .... + a_{125} = {4.4}^{20} = 2^{41}$
and $a_{125} = 2^{20} \Rightarrow a_{1} + a_{3} + .... + a_{123} = 2^{41} - 2^{20}$
on comparing the coefficient
$a_{0} = 0, a_{1} = 0, a_{2} = 0, a_{3} = 1$