The expression to evaluate is $\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$, under the condition that $k^{101}=1$ and $k\ne1$. A participant suggests that the upper limit should be 100 instead of 200, leading to a reevaluation of the series. The final calculation results in a value of 49, derived from the sum of terms and the evaluation of the series involving $\dfrac{1}{k^n-1}$. The discussion highlights the importance of careful notation and the impact of minor errors on the overall solution.
#1
anemone
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If $k^{101}=1$ and $k\ne1$, evaluate $\dfrac{k^2}{k-1}+\dfrac{k^4}{k^2-1}+\dfrac{k^6}{k^3-1}+\cdots+\dfrac{k^{200}}{k^{100}-1}$.
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
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#5
kaliprasad
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anemone said:
Thanks for participating, Albert!
But I think there is a typo, where I think the 200 is actually a 100:
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 $k-1 + \dfrac{1}{k-1}+ 2$
#6
anemone
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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!
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