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Question

If $${(1+i)}^{2n}+{(1-i)}^{2n}=-{2}^{n+1}$$ where, $$i=\sqrt{-1}$$ for all those $$n$$, which are


A
even
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B
odd
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C
multiple of 3
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D
None of these
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Solution

The correct option is A even
In $$(1+i)^{2n}+(1-i)^{2n}$$
$$=\left\{(1+i)^{2}\right\}^{n}+\left\{(1-i)^{2}\right\}^n$$
$$=(1+i^{2}+2i) ^{n}+(1+i^{2}-2i) ^{n}$$
$$=(1-1+2\ i)^{n}+(1-1-2\ i) ^{n}$$
$$=2^{n}i^{2}+i^{2}(-2) ^n$$
$$=i^{2}(2^{2}+(-2) ^n)$$
When $$n=2$$
$$= i^{2}(2^{2}+2^{2})=-1 \cdot 2^{2+1}$$
Hence, $$’n’$$ must be even

Mathematics

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