If \rho \neq 1 is an nth root of unity, then 1+5 \rho+9 \rho^{2}+. . . . .+(4 n-3) \rho^{2-1}=\frac{\text{ an }}{\rho-1} and deduce the values of the following. 1+5 \cos (\frac{2 \pi}{ {n}})+9 \cos (\frac{4 \pi}{ {n}})+. . . . .+(4 {n}-3) \cos ( {2( {n}-1) \pi}/{ {n}})= {bn} 5 \sin (\frac{2 \pi}{n})+9 \sin (\frac{4 \pi}{n})+. . . .+(4 n-3) \sin (\frac{2(n-1) \pi}{n})=con \cot (\frac{\pi}{n}) for n \geq 2, then find the value of a-b-c.

JEE Advanced 2020 Concept Recapitulation Test 3 Paper 2

Question : 38 of 54

Marks: +1, -0

If ρ≠1 is an nth root of unity, then 1+5ρ+9ρ2+.....+(4n−3)ρ2−1=‌

‌ an ‌

ρ−1

and deduce the values of the following. 1+5‌cos(‌

2π

)+9‌cos(‌

4π

)+.....+(4n−3)‌cos(‌

2(n−1)π

)=bn 5‌sin(‌

2π

)+9‌sin(‌

4π

)+....+(4n−3)‌sin(‌

2(n−1)π

)=con‌cot(‌

) for n≥2, then find the value of a−b−c.

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