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Observe that we can find the sum $\sum\limits_{i=1}^{n}i^3$ using "integration": $$\sum\limits_{i=1}^{n}\frac{i^3}{3}=\int\sum\limits_{i=1}^{n}i^2dn.$$ and actually we get exactly the right result(if we know the sum of $i^2$ in terms of $n$) For other $k$ this approach is not generally true, but by integrating the sum we still get the right coefficient. What is the reason behind this method? Does this integration have a name and can we utilize the method to get the sum $\sum\limits_{i=1}^{n}i^k$ in terms of $n$?

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    $\begingroup$ The integral of $i^2$ is not $i^3/3$ because here $i$ is a constant. It doesn't even make sense to say $di$ when $i$ is a fixed integer like that. $\endgroup$ – Gregory Grant Nov 13 '17 at 20:42
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Let $S_k(n)=\sum_{i=1}^n i^k$. It's easy to prove that $S_k(n)$ is a polynomial of degree $k+1$ in $n$. It is characterised by $S_k(0)=0$ and $S_k(x)-S_k(x-1)=x^k$. Differentiating the last gives $$S_k'(x)-S_k'(x-1)=kx^{k-1}=k(S_{k-1}(x)-S_{k-1}(x-1)).$$ Therefore $$S_k'(x)-kS_{k-1}(x)=S_k'(x-1)-kS_{k-1}(x-1)$$ which implies that $S'_k(x)-kS_{k-1}(x)$ is a constant. Let's call that $C_k$. Then $$S_k(x)=\int_0^x(kS_{k-1}(t)+C_k)\,dt =C_kx+k\int_0^x S_{k-1}(t)\,dt.$$ Of course, $C_k$ is determined by the condition that $S_k(1)=1$. So we more-or-less get $S_k$ by integrating $kS_{k-1}$....

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\begin{eqnarray*} \sum_{i=1}^{n} i^4 &=& \frac{1}{5} n^5 &+& \frac{1}{2} n^4 &+& \frac{1}{3}n^3 &-& \frac{1}{30}n \\ \sum_{i=1}^{n} i^5 &=& \frac{1}{6} n^6 &+& \frac{1}{2} n^5 &+& \frac{5}{12}n^4 &-& \frac{1}{12}n^2 \\ \sum_{i=1}^{n} i^6 &=& \frac{1}{7} n^7 &+& \frac{1}{2} n^6 &+& \frac{1}{2}n^5 &-& \frac{1}{6}n^3+ \frac{1}{42}n \\ \end{eqnarray*} This works in general when going from the even to the next odd exponent but when going from the odd to even another term is introduced. This can be seen from Faulhauber's formula and the observation that the "odd" Bernoulli numbers are zero. See here for more about Faulhauber's formula https://en.wikipedia.org/wiki/Faulhaber%27s_formula

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