Calculate $\sum_{i=0}^ni^k$ where $k$ is given, $k\in\mathbb{N}$.

Question

Calculate $$\sum_{i=0}^n i^k$$ where $k$ is given and belongs to $\mathbb{N}$.

Answer 1

For a relatively easy solution by hand, consider

$$n^2=\sum_{i=1}^ni^2-\sum_{i=1}^n(i-1)^2=2\sum_{i=1}^ni-\sum_{i=1}^n1=2S_1-S_0$$ so that

$$S_1=\frac{n^2+S_0}2=\frac{n^2+n}2.$$

Then

$$n^3=\sum_{i=1}^ni^3-\sum_{i=1}^n(i-1)^3=3S_2-3S_1+S_0$$ so that

$$S_2=\frac{n^3+3S_1-S_0}3=\frac{2n^3+3n^2+n}6.$$

Next

$$S_3=\frac{n^4+6S_2-4S_1+S_0}4$$ $$\cdots$$

You can continue at will, using the Binomial theorem. By substituting the $S_k$, you obtain explicit polynomial forms. There is a general expression, involving the Bernouilli numbers.

Answer 2

We construct the series $(x_n)_{n\ge 0}$ as follows: $x_i = (i+1)^k+1,k\ge 0$. We then have:

$$x_i=(i+1)^k=i^{k+1}+\binom{1}{k+1}i^k+\binom{2}{k+1}i^{k-1}+...+\binom{k}{k+1}i+1$$

For $i=1$:

$2^{k+1}=1+\binom{1}{k+1}+\binom{2}{k+1}+...+\binom{k}{k+1}+1$

For $i=2$:

$3^{k+1}=2^{k+1}+\binom{1}{k+1}2^k+\binom{2}{k+1}2^{k-1}+...+\binom{k}{k+1}2+1$

For $i=3$:

$4^{k+1}=3^{k+1}+\binom{1}{k+1}3^k+\binom{2}{k+1}3^{k-1}+...+\binom{k}{k+1}3+1$

$$\cdots$$

For $i=n+1$:

$(n+1)^k=n^{k+1}+\binom{1}{k+1}n^k+\binom{2}{k+1}n^{k-1}+...+\binom{k}{k+1}n+1$.

We sum up all the above equalities and we get:

$$\sum_{i=0}^n(i+1)^{k+1}=\sum_{i=0}^{n-1}(i+1)^{k+1}+\binom{1}{k+1}\sum_{i=0}^{n-1}(i+1)^{k}+\binom{2}{k+1}\sum_{i=0}^{n-1}(i+1)^{k-1}+\cdots+\binom{k}{k+1}\sum_{i=0}^{n-1}(i+1)+1$$

By moving $\sum_{i=0}^{n-1}(i+1)^{k+1}$ to the left-hand side we have:

$$(n+1)^{k+1}=\binom{1}{k+1}\sum_{i=0}^{n-1}(i+1)^{k}+\binom{2}{k+1}\sum_{i=0}^{n-1}(i+1)^{k-1}+\cdots+\binom{k}{k+1}\sum_{i=0}^{n-1}(i+1)+1$$