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Question

If t2+t+1=0, then (t+1t)2+(t2+1t2)2+(t3+1t3)2....+(t27+1t27)2 is equal to

A
54
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B
27
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C
18
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D
0
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Solution

The correct option is A 54
t2+t+1=0
t=1+i32,1i32=w,w2
Let t=w1t=1w=w2
t+1t=w+w2=1
If r is not divisile by 3,then
tr+1tr=wr+(w2)r=wr+w2r=1
If r is divisible by 3,then
tr+1tr=2
Hence (t+1t)2+(t2+1t2)2+(t3+1t3)2....+(t27+1t27)2
=(1+1+1+....18 times)+(4+4+4+....9 times)=54

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