Dear Student,Prove that: 12-α1+12-α2+ +12-αn-1=(n-2)2n-1+12n-1Now, We know that,If 1, α1,α2, αn are nth root of unity, then;1x-1+1x-α+1x-α2+ +1x-αn-1=nxn-1xn-1Now put x=2, we get;⇒12-1+12-α1+12-α2+ +12-αn-1=n2n-12n-1⇒1+12-α1+12-α2+ +12-αn-1=n2n-12n-1⇒12-α1+12-α2+ +12-αn-1=n2n-12n-1-1⇒12-α1+12-α2+ +12-αn-1=(n×2n×2-1)-(2n-1)2n-1⇒12-α1+12-α2+ +12-αn-1=(n×2n×2-1)-2n+12n-1⇒12-α1+12-α2+ +12-αn-1=2n((n×2-1)-1)+12n-1⇒12-α1+12-α2+ +12-αn-1=2n(n2-1)+12n-1⇒12-α1+12-α2+ +12-αn-1=2n2(n-2)+12n-1⇒12-α1+12-α2+ +12-αn-1=(n-2)2n-1+12n-1; Hence ProvedRegards

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Solve this: ... an _ are the nth roots of unity, then prove

Dear Student, Prove that : \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{(n - 2) 2^{n - 1} + 1}{2^{n} - 1} Now, We know that, If 1, α_{1}, α_{2}, . . . . . . . α_{n} are nth root of unity, then; \frac{1}{x - 1} + \frac{1}{x - α} + \frac{1}{x - α_{2}} + . . . . . + \frac{1}{x - α_{n - 1}} = \frac{{nx}^{n - 1}}{x^{n} - 1} Now put x = 2, we get; \Rightarrow \frac{1}{2 - 1} + \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{n 2^{n - 1}}{2^{n} - 1} \Rightarrow 1 + \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{n 2^{n - 1}}{2^{n} - 1} \Rightarrow \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{n 2^{n - 1}}{2^{n} - 1} - 1 \Rightarrow \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{(n \times 2^{n} \times 2^{- 1}) - (2^{n} - 1)}{2^{n} - 1} \Rightarrow \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{(n \times 2^{n} \times 2^{- 1}) - 2^{n} + 1}{2^{n} - 1} \Rightarrow \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{2^{n} ((n \times 2^{- 1}) - 1) + 1}{2^{n} - 1} \Rightarrow \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{2^{n} (\frac{n}{2} - 1) + 1}{2^{n} - 1} \Rightarrow \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{\frac{2^{n}}{2} (n - 2) + 1}{2^{n} - 1} \Rightarrow \frac{1}{2 - α_{1}} + \frac{1}{2 - α_{2}} + . . . . . + \frac{1}{2 - α_{n - 1}} = \frac{(n - 2) 2^{n - 1} + 1}{2^{n} - 1}; Hence Proved Regards .

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Solve this: Solve this: ... an _ are the nth roots of unity, then prove

Solve this:

Solve this: ... an _ are the nth roots of unity, then prove