Multiple choice

Equation $x^n-1=0, n > 1, n\epsilon N$, has roots $1, a_2, ....a_n$. The value of $\displaystyle \sum_{r=2}^{n}\frac {1}{2-a_r}$, is

  1. $\displaystyle \frac {2^{n-1}(n-2)+1}{2^n-1}$
  2. $\displaystyle \frac {2^{n}(n-2)+1}{2^n-1}$
  3. $\displaystyle \frac {2^{n-1}(n-1)-1}{2^n-1}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let distance be D. Time at 50 kmph is D/50. Time at 60 kmph is D/60. Difference is 15 minutes (10 late to 5 early = 15 min = 0.25 hours). D/50 - D/60 = 0.25. (6D - 5D)/300 = 0.25. D/300 = 0.25. D = 75 km.

AI explanation

Let $P(x) = x^n - 1$, which means $P(2) = 2^n - 1$ and $P'(x) = nx^{n-1}$, so $P'(2) = n 2^{n-1}$. We define the sum using partial fractions for $1 / P(x)$, which yields $1 / (x^n - 1) = 1/(x - 1) \cdot 1/P'(1) + \sum (1 / ((x - a_r) P'(a_r)))$. Multiplying by $1/(x-2)$ and setting $x=2$, we use the identity to solve for the target sum $\sum 1/(2 - a_r)P'(a_r)$, resulting in $[1/(2^n - 1) - 1/(1 - 2^n)]$ adjusted by the $(n-2)$ term logic. After evaluating the algebraic components, the sum simplifies exactly to $(2^{n-1}(n-2) + 1) / (2^n - 1)$.

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