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Another way (by Euler, I think), starting from the geometric sum:

$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$

Differentiate both sides:

$$1 + 2 x + 3 x^2 + \cdots + n x^{n-1} = \frac{n x^{n+1}-(n+1) x^{n} +1}{(x-1)^2} \tag{2}$$

Multiply by $x$ and differentiate once more. We get on the LHS

$$1 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$

which, evaluated at $x=1$ gives our desired sum $\sum_{k=1}^n k^2$. Hence, we just need to multiply by $x$ the RHS of $(2)$, compute the (eg, with L'Hopital rule) and evaluate it at $x \to 1$.

It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.

(Update) here's a concrete computation, applying the binomial theorem on the RHS of $(1)$ and doing a series expansion around $x=1$. Let

$$\begin{align} g(x)&=\frac{x^{n+1}-1}{x-1}\\ &=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\ &={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4} \end{align}$$

Differentiating, multiplying by $x$ and differentiating again we get $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$ ... which evaluated at $x=1$ gives the desired answer: $${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6} $$

We can apply the same procedure for higher powers. For example:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$

Another way (by Euler, I think), starting from the geometric sum:

$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$

Differentiate both sides:

$$1 + 2 x + 3 x^2 + \cdots + n x^{n-1} = \frac{n x^{n+1}-(n+1) x^{n} +1}{(x-1)^2} \tag{2}$$

Multiply by $x$ and differentiate once more. We get on the LHS

$$1 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$

which, evaluated at $x=1$ gives our desired sum $\sum_{k=1}^n k^2$. Hence, we just need to multiply by $x$ the RHS of $(2)$, compute the derivative and evaluate the limit at $x \to 1$.

It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.

(Update) here's a concrete computation, applying the binomial theorem on the RHS of $(1)$ and doing a series expansion around $x=1$. Let

$$\begin{align} g(x)&=\frac{x^{n+1}-1}{x-1}\\ &=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\ &={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4} \end{align}$$

Differentiating, multiplying by $x$ and differentiating again we get $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$ ... which evaluated at $x=1$ gives the desired answer: $${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6} $$

We can apply the same procedure for higher powers. For example:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$

Another way (by Euler, I think), from the geometric sum:

$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$

Differentiate both sides and multiply by $x$:

$$x + 2 x^2 + 3 x^3 + \cdots + n x^{n} = \frac{n x^{n+2}-(n+1) x^{n+1} +x}{(x-1)^2} \tag{2}$$

Differentiating once more, we get on the LHS

$$1 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$

which, evaluated at $x=1$ gives our desired sum $\sum_{k=1}^n k^2$. Hence, we just need to compute the derivative on the RHS in $(2)$  (eg, with L'Hopital rule) and evaluate it at $x \to 1$.

It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.

(Update) here's a concrete computation, applying the binomial theorem on the RHS of $(1)$ and doing a series expansion around $x=1$. Let

$$\begin{align} g(x)&=\frac{x^{n+1}-1}{x-1}\\ &=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\ &={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4} \end{align}$$

Deriving, multiplying by $x$ and deriving again: $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$ which evaluated at $x=1$ gives the desired answer: $${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6} $$

We can apply the same procedure for higher powers. For example:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$

Another way (by Euler, I think), starting from the geometric sum:

$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$

Differentiate both sides:

$$1 + 2 x + 3 x^2 + \cdots + n x^{n-1} = \frac{n x^{n+1}-(n+1) x^{n} +1}{(x-1)^2} \tag{2}$$

Multiply by $x$ and differentiate once more. We get on the LHS

$$1 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$

which, evaluated at $x=1$ gives our desired sum $\sum_{k=1}^n k^2$. Hence, we just need to multiply by $x$ the RHS of $(2)$, compute the (eg, with L'Hopital rule) and evaluate it at $x \to 1$.

It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.

(Update) here's a concrete computation, applying the binomial theorem on the RHS of $(1)$ and doing a series expansion around $x=1$. Let

$$\begin{align} g(x)&=\frac{x^{n+1}-1}{x-1}\\ &=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\ &={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4} \end{align}$$

Differentiating, multiplying by $x$ and differentiating again we get $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$ ... which evaluated at $x=1$ gives the desired answer: $${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6} $$

We can apply the same procedure for higher powers. For example:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$

Another way (by Euler, I think), from the geometric sum:

$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$

Differentiate both sides and multiply by $x$:

$$x + 2 x^2 + 3 x^3 + \cdots + n x^{n} = \frac{n x^{n+2}-(n+1) x^{n+1} +x}{(x-1)^2} \tag{2}$$

Differentiating once more, we get on the LHS

$$1 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$

which, evaluated at $x=1$ gives our desired sum $\sum_{k=1}^n k^2$. Hence, we just need to compute the derivative on the RHS in $(2)$ (eg, with L'Hopital rule) and evaluate it at $x \to 1$.

It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.

(Update) here's a concrete computation, by doing a series expansion of around $(x-1)$, applying the binomial theorem on the RHS of $(1)$. Let

$$\begin{align} g(x)&=\frac{x^{n+1}-1}{x-1}\\ &=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\ &={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4} \end{align}$$

Deriving, multiplying by $x$ and deriving again: $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$ which evaluated at $x=1$ gives the desired answer: $${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6} $$

We can apply the same procedure for higher powers. For example:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$

Another way (by Euler, I think), from the geometric sum:

$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$

Differentiate both sides and multiply by $x$:

$$x + 2 x^2 + 3 x^3 + \cdots + n x^{n} = \frac{n x^{n+2}-(n+1) x^{n+1} +x}{(x-1)^2} \tag{2}$$

Differentiating once more, we get on the LHS

$$1 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$

which, evaluated at $x=1$ gives our desired sum $\sum_{k=1}^n k^2$. Hence, we just need to compute the derivative on the RHS in $(2)$ (eg, with L'Hopital rule) and evaluate it at $x \to 1$.

It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.

(Update) here's a concrete computation, applying the binomial theorem on the RHS of $(1)$ and doing a series expansion around $x=1$. Let

$$\begin{align} g(x)&=\frac{x^{n+1}-1}{x-1}\\ &=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\ &={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4} \end{align}$$

Deriving, multiplying by $x$ and deriving again: $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$ which evaluated at $x=1$ gives the desired answer: $${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6} $$

We can apply the same procedure for higher powers. For example:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$