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Question

If $${(1+x+2{x}^{2})}^{20}={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+.....+{a}_{10}{x}^{40}$$, then $${a}_{1}+{a}_{3}+{a}_{5}+......+{a}_{39}$$ equals


A
219(2201)
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B
220(21919)
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C
219(220+21)
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D
None of these
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Solution

The correct option is A $${2}^{19}({2}^{20}-1)$$
$$(1+x+2x^{2})^{20}=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+.......+a_{40}x^{40}$$
$$x=1$$
$$(1+1+2)^{20}=a_{0}+a_{1}+a_{2}+a_{3}+....+a_{40}$$ ________ (1)
$$x=-1$$
$$(1-1+2)^{20}=a_{0}-a_{1}+a_{2}-a_{3}+.....+a_{40}$$_________(2)
(1) _ (2)
$$4^{20}-2^{20}=2(a_{1}+a_{3}+........+a_{39})$$
$$2^{20}(2^{20}-1)=2(a_{1}+a_{3}+......+a_{39})$$
$$a_{1}+a_{3}+......+a_{39}=2^{19}(2^{20}-1)$$

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