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$$ \sum_{k=1}^{n} k^r = \frac{1}{r+1} n^{r+1} + a_{r,r} n^r + \ldots + a_{r,1} n $$ I'm apparently supposed to use the binomial theorem, but I can't seem to figure it out. The note says: "For the induction step, first consider the term $(k-1)^{r+1}\$ and cleverly transform it using the binomial theorem."

I'm really having trouble with the base case and the induction step. I also saw the post: $\sum\limits_{k=1}^nk^r = \frac{1}{1+r}n^{r+1}+a_rn^r +....+a_1n$ but I still don't understand it.

Arthur
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Kadir
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  • As you noticed, this has been asked and answered before. Please be more specific about what you do not understand. – Martin R Nov 06 '23 at 08:22
  • See this answer. In effect, what you are focusing on is an algebra-only (i.e. no calculus) derivation, which can be considered the 1st step in an algebra-only derivation of bernoulli numbers. The linked answer itself contains a second link, which describes how to solve your problem in the specific case of $~r = 3.~$ ...see next comment – user2661923 Nov 06 '23 at 08:48
  • So, you solve the problem by considering the general case of $~k^r,~$ and deriving the $~(r+1)~$ linear equations in $~(r+1)~$ unknowns. – user2661923 Nov 06 '23 at 08:50
  • @MartinR The link in my question does not show the base case for n=1 and I don't know how to interpret a_{r,r}. I know the left side of the equation is equal to 1 but right side is giving me trouble. As for the induction step: the link in my question says at the end of the top answer "Upon expanding the terms of: ..." but the answer does not show the same thing as in the induction hypothesis. So my question is how all those terms with (n+1-1)^r can be transformed in a way that it looks the same as in the induction hypothesis. – Kadir Nov 06 '23 at 09:32
  • @user2661923 Why does the link in my question then prove it for all n ∈ N? Also we did not have these kinds of methods in my real analysis course so far. Is there a simpler way you can show it or could you kindly write this proof out for me? Because I thought that r can be any number? Why would this proof be correct if I show it for r=3? – Kadir Nov 06 '23 at 09:42
  • To Kadir: Why not ignore your work and simply focus on the link that I provided, which links to a second article? I suspect that if you do that, you will learn how to validly attack the problem. Then, you can compare my method with your original method. – user2661923 Nov 06 '23 at 12:20
  • To Kadir: "Why would this proof be correct if I show it for r=3?" : It wouldn't be correct. What you have to do is study the $~r=3~$ proof to the point that you have a complete understanding of it. Then, you would have to generalize it, showing that very similar analysis will apply for any $~r \in \Bbb{Z^+}.$ – user2661923 Nov 06 '23 at 20:50

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