If `((2n+1),(0))+ ((2n+1),(3)) +((2n+1),(6))+ ...= 170, ` then n
equals

Options

A. 2
B. 4
C. 6
D. 8

Answer

b

Explanation

`because (1+x)^(2n+1)=""^(2n+1)C_(0) + ""^(2n+1)C_(1)x +""^(2n+1)C_(2)x^(2)+""^(2n+1)C_(3)x^(3)`
` + ""^(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`