If for z as real or complex, 1+z2+z48=C0+C1z2+C2z4+⋯+C16z32, then

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C0−C1+C2−C3+⋯+C16=1

C0+C3+C6+C9+C12+C15=37

C2+C5+C8+C11+C14=36

C1+C4+C7+C10+C13+C16=37

answer is A.

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Detailed Solution

1+z2+z48=C0+C1z2+C2z4+⋯+C16z32 (1)Putting z = i, where i=−1,(1−1+1)8=C0−C1+C2−C3+⋯+C16or C0−C1+C2−C3+⋯+C16=1Also, putting z=ω,1+ω2+ω48=C0+C1ω2+C2ω4+⋯+C16ω32or C0+C1ω2+C2ω+C3+⋯+C16ω2=0 (2)Putting x=ω2,1+ω4+ω88=C0+C1ω4+C2ω8+⋯+C16ω64or C0+C1ω+C2ω2+⋯+C16ω=0 (3)Putting x = 1,38=C0+C1+C2+⋯+C16 (4)Adding (2), (3) and (4), we have3C0+C3+⋯+C15=38or C0+C3+⋯+C15=37Similarly, first multiplying (1) by z and then putting 1, ω, ω2 and adding, we getC1+C4+C7+C10+C13+C16=37Multiplying (1) by z2 and then putting 1, ω, ω2 and adding, we getC2+C5+C8+C11+C14=37

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