Question: Use the method of Problem 6 to show that $\sum_{i=0}^n k^p$ can always be written in the form $\frac{n^{p+1}}{p+1} +An^p + Bn^{p-1} + Bn^{p-2} + Cn^{p-3} + ..... $

The method in 6 they are talking about is the telescoping method.

I have tried to derive the solution for some while and I somewhat came to a proof (there's so much constant terms I am kind of confused). The solution from the manual is: enter image description here enter image description here

I can get a sense of the proof, but I am not quite sure how exactly the part after --

"Adding for $k=1,...,n$, we obtain" to the end is exactly formed. An explanation would be very helpful.



If you aren't familiar with the notation, when I write $\mathcal{O}(k^r)$, I basically mean "terms involving $k^r$". Then it should be clear that $\mathcal{O}(k^r) = \frac{1}{p+1} \mathcal{O}(k^r)$.

In this case, we have $r < p$, so we have

$$(k+1)^{p+1} - k^{p+1} = (p+1)k^p + \mathcal{O}(k^r),$$ so $$\frac{(k+1)^{p+1} - k^{p+1}}{p+1} = k^p + \frac{\mathcal{O}(k^r)}{p+1} = k^p + \mathcal{O}(k^r),$$ then you can write $$ \sum_{k=1}^n k^p + \sum_{k=1}^n \mathcal{O}(k^r)= \sum_{k=1}^n \frac{(k+1)^{p+1} - k^{p+1}}{p+1} = \frac{(n+1)^{p+1}}{p+1} - \frac{1}{p+1}, $$ then you can add $\frac{1}{p+1}$ over to the left side and absorb it into $\mathcal{O}(k^r)$ (since $1 = k^0$).

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  • $\begingroup$ And what about the terms involving the lower powers of $k$, $k^r$ ($r<p$) in the Right hand side of the equation? Doesn't their co-efficients get divided by $(p+1)$ as well? $\endgroup$ – Rio1210 Mar 22 '17 at 21:03
  • $\begingroup$ I missed that part actually. Let me see if I can fix things. $\endgroup$ – amarney Mar 22 '17 at 21:12
  • $\begingroup$ @amarney: I think that it should be $\dfrac{(n+1)^{p+1}}{p+1}-\dfrac{1}{p+1}$ instead. $\endgroup$ – K.M Jun 27 '19 at 2:33
  • $\begingroup$ yup, forgot that, fixed $\endgroup$ – amarney Jun 28 '19 at 13:18

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