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If ((2n+1),(0))+ ((2n+1),(3)) +((2n+1),(6))+ ...= 170,then n equals |
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Answer» 2 ` + ""^(2n+1)C_(4)x^(4)+""^(2n+1)C_(5)x^(5)+""^(2n+1)C_(6)x^(6)+...` Putting `x=1, omega, omega^(2)` (where `omega` is cube root of unity and adding, we GET `2^(2n+1)+(1+omega) ""^(2n+1)+(1+omega^(2))^(2n+1)=3(""^(2n+1)C_(0) +""^(2n+1)C_(3)+""^(2n+1)C_(6)+...)` `rArr 2""^(2n+1)-omega ^(2) " "^(2n+1)- omega^(2n+1)=3(""^(2n+1)C_(0) +""^(2n+1)C_(3)+""^(2n+1)C_(6)+...)[because 1 + omega + omega^(2) = 0]` `rArr ""^(2n+1)C_(0) +^(2n+1)C_(3)+^(2n+1)C_(6)+...=1/3 ` `(2^(2n+1) -omega^(2) " "^(2n+1) -omega^(2n+1))` `((2n+1),(0))+ ((2n+1),(3)) +((2n+1),(6))+...=1/3(2^(2n+1) -omega^(2) " "^(2n+1) -omega^(2n+1))` For N `= 4,170 = 1/3 (512-1-1)=170 [because omega^(2)=1]` Hence,`n=4` |
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