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

[anemone]

If $k^{101}=1$ and $k\ne1$, evaluate $\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$.

 

[Albert1]

If $k^{101}=1$ and $k\ne1$, evaluate $\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$.

Spoiler

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

 

[anemone]

Spoiler

$\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$
$=(k-1)+\dfrac{1}{k-1}+2+(k^2-1)+\dfrac{1}{k^2-1}+2+--+(k^{100}-1)+\dfrac{1}{k^{100}-1}+2$
$=(k+k^2+k^3+---+k^{100})+200+(\dfrac{1}{k-1}+\dfrac{1}{k^2-1}+---+\dfrac{1}{k^{100}-1})$
$=-1+200-50=149$

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 $-50$:

Spoiler

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

 

[Albert1]

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

 

[kaliprasad]

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 $-50$:

Spoiler

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

Spoiler

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 $k-1 + \dfrac{1}{k-1}+ 2$

 

[anemone]

Thanks! yes it is " 100"

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

$\dfrac{1}{k-1}+\dfrac {1}{k^{100}-1}$
$=\dfrac{1}{k-1}+\dfrac{k}{1-k}=\dfrac{k-1}{1-k}=-1$
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!