Evaluating an Expression with $k^{101}=1$ and $k\ne1$

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Discussion Overview

The discussion revolves around evaluating the expression $\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$ under the condition that $k^{101}=1$ and $k\ne1$. Participants explore the implications of this expression, including potential typographical errors and the reasoning behind their evaluations.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest that there is a typo in the original expression, proposing that the upper limit should be 100 instead of 200.
  • One participant evaluates the expression and arrives at a total of 49, breaking down the components into sums involving $k$ and fractions.
  • Another participant elaborates on the sum of the fractions, claiming it yields -50 based on pairing terms.
  • There is a discussion about the evaluation of $\dfrac{1}{k-1}+\dfrac{1}{k^{100}-1}$ leading to a conclusion of -1, which is part of the reasoning for the overall sum.
  • Participants express a mix of confidence and self-reflection regarding their evaluations, with some acknowledging mistakes in their calculations.

Areas of Agreement / Disagreement

Participants generally agree on the need to correct the upper limit of the expression from 200 to 100. However, there is no consensus on the final evaluation of the expression, as different interpretations and calculations are presented.

Contextual Notes

There are unresolved assumptions regarding the manipulation of the series and the properties of $k$ under the given conditions. The discussion includes potential errors in calculations and interpretations that have not been definitively resolved.

anemone
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If $k^{101}=1$ and $k\ne1$, evaluate $\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$.
 
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anemone said:
If $k^{101}=1$ and $k\ne1$, evaluate $\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$.
$\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$
$=(k-1)+\dfrac{1}{k-1}+2+(k^2-1)+\dfrac{1}{k^2-1}+2+--+(k^{100}-1)+\dfrac{1}{k^{100}-1}+2$
$=(k+k^2+k^3+---+k^{100})+100+(\dfrac{1}{k-1}+\dfrac{1}{k^2-1}+---+\dfrac{1}{k^{100}-1})$
$=-1+100-50=49$
 
Last edited:
Albert said:
$\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$
$=(k-1)+\dfrac{1}{k-1}+2+(k^2-1)+\dfrac{1}{k^2-1}+2+--+(k^{100}-1)+\dfrac{1}{k^{100}-1}+2$
$=(k+k^2+k^3+---+k^{100})+200+(\dfrac{1}{k-1}+\dfrac{1}{k^2-1}+---+\dfrac{1}{k^{100}-1})$
$=-1+200-50=149$

Thanks for participating, Albert!

But I think there is a typo, where I think the 200 is actually a 100:

$\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$
$=(k-1)+\dfrac{1}{k-1}+2+(k^2-1)+\dfrac{1}{k^2-1}+2+--+(k^{100}-1)+\dfrac{1}{k^{100}-1}+2$
$=(k+k^2+k^3+---+k^{100})+{\color{red}{100}}+(\dfrac{1}{k-1}+\dfrac{1}{k^2-1}+---+\dfrac{1}{k^{100}-1})$
$=-1+{\color{red}{100}}-50=49$

Just to further elaborate, based on my solution, about how the expression $\dfrac{1}{k-1}+\dfrac{1}{k^2-1}+---+\dfrac{1}{k^{100}-1}$ yields a $-50$:

$\begin{align*}\dfrac{1}{k-1}+\dfrac{1}{k^2-1}+---+\dfrac{1}{k^{100}-1}&=\left(\dfrac{1}{k-1}+\dfrac{1}{k^{100}-1}\right)+\left(\dfrac{1}{k^2-1}+\dfrac{1}{k^{99}-1}\right)+\cdots+\left(\dfrac{1}{k^{50}-1}+\dfrac{1}{k^{51}-1}\right)\\&=\left(\dfrac{(k^{100}-1)+(k-1)}{(k-1)(k^{100}-1)}\right)+\left(\dfrac{(k^{99}-1)+(k^2-1)}{(k^2-1)(k^{99}-1)}\right)+\cdots+\left(\dfrac{(k^{51}-1)+(k^{50}-1)}{(k^{50}-1)(k^{51}-1)}\right)\\&=\left(\dfrac{k^{100}+k-2}{k^{101}-k-k^{100}-1}\right)+\left(\dfrac{k^{99}+k^2-2}{k^{101}-k^2-k^{99}+1}\right)+\cdots+\left(\dfrac{k^{51}+k^{50}-2}{k^{101}-k^{50}-k^{51}+1}\right)\\&=\left(\dfrac{k^{100}+k-2}{-(k^{100}+k-2)}\right)+\left(\dfrac{k^{99}+k^2-2}{-(k^{99}+k^2-2)}\right)+\cdots+\left(\dfrac{k^{51}+k^{50}-2}{-(k^{51}+k^{50}-2)}\right)\\&=-50\end{align*}$
 
Thanks! yes it is " 100"
$\dfrac{1}{k-1}+\dfrac {1}{k^{100}-1}$
$=\dfrac{1}{k-1}+\dfrac{k}{1-k}=\dfrac{k-1}{1-k}=-1$
for $k^{100}=\dfrac {1}{k},\,\,\, \, k^{99}=\dfrac {1}{k^2},------$
we have 50 paires ,it sums up to -50
 
Last edited:
anemone said:
Thanks for participating, Albert!

But I think there is a typo, where I think the 200 is actually a 100:

$\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$
$=(k-1)+\dfrac{1}{k-1}+2+(k^2-1)+\dfrac{1}{k^2-1}+2+--+(k^{100}-1)+\dfrac{1}{k^{100}-1}+2$
$=(k+k^2+k^3+---+k^{100})+{\color{red}{100}}+(\dfrac{1}{k-1}+\dfrac{1}{k^2-1}+---+\dfrac{1}{k^{100}-1})$
$=-1+{\color{red}{100}}-50=49$

Just to further elaborate, based on my solution, about how the expression $\dfrac{1}{k-1}+\dfrac{1}{k^2-1}+---+\dfrac{1}{k^{100}-1}$ yields a $-50$:

$\begin{align*}\dfrac{1}{k-1}+\dfrac{1}{k^2-1}+---+\dfrac{1}{k^{100}-1}&=\left(\dfrac{1}{k-1}+\dfrac{1}{k^{100}-1}\right)+\left(\dfrac{1}{k^2-1}+\dfrac{1}{k^{99}-1}\right)+\cdots+\left(\dfrac{1}{k^{50}-1}+\dfrac{1}{k^{51}-1}\right)\\&=\left(\dfrac{(k^{100}-1)+(k-1)}{(k-1)(k^{100}-1)}\right)+\left(\dfrac{(k^{99}-1)+(k^2-1)}{(k^2-1)(k^{99}-1)}\right)+\cdots+\left(\dfrac{(k^{51}-1)+(k^{50}-1)}{(k^{50}-1)(k^{51}-1)}\right)\\&=\left(\dfrac{k^{100}+k-2}{k^{101}-k-k^{100}-1}\right)+\left(\dfrac{k^{99}+k^2-2}{k^{101}-k^2-k^{99}+1}\right)+\cdots+\left(\dfrac{k^{51}+k^{50}-2}{k^{101}-k^{50}-k^{51}+1}\right)\\&=\left(\dfrac{k^{100}+k-2}{-(k^{100}+k-2)}\right)+\left(\dfrac{k^{99}+k^2-2}{-(k^{99}+k^2-2)}\right)+\cdots+\left(\dfrac{k^{51}+k^{50}-2}{-(k^{51}+k^{50}-2)}\right)\\&=-50\end{align*}$

we have
$\dfrac{k^2}{k-1}$
= $\dfrac{k^2-1+1}{k-1}$
= $k+1 + \dfrac{1}{k-1}$

which is straight forward

why $k-1 + \dfrac{1}{k-1}+ 2$
 
Albert said:
Thanks! yes it is " 100"

See, I told you so...hehehe...:p but I know that was purely an honest mistake.

Albert said:
$\dfrac{1}{k-1}+\dfrac {1}{k^{100}-1}$
$=\dfrac{1}{k-1}+\dfrac{k}{1-k}=\dfrac{k-1}{1-k}=-1$
for $k^{100}=\dfrac {1}{k},\,\,\, \, k^{99}=\dfrac {1}{k^2},------$
we have 50 paires ,it sums up to -50

Argh! How could I miss out something so "obvious" like that? Shame on me!:mad:
 

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