The sum $$\sum_{k=1}^{20} (1 + 2 + 3 + \ldots + k)$$ is

We have to evaluate the series $$\displaystyle\sum_{k=1}^{20}\left(1+2+3+\dots+k\right).$$

First recall the standard result for the sum of the first $$k$$ natural numbers. The formula is

$$1+2+3+\dots+k=\frac{k(k+1)}{2}.$$

Using this, the given series becomes

$$\sum_{k=1}^{20}\left(1+2+3+\dots+k\right)=\sum_{k=1}^{20}\frac{k(k+1)}{2}.$$

Because the constant $$\tfrac12$$ can be factored out of the summation, we write

$$\sum_{k=1}^{20}\frac{k(k+1)}{2}=\frac12\sum_{k=1}^{20}k(k+1).$$

Now expand $$k(k+1)$$:

$$k(k+1)=k^2+k.$$

Substituting this expansion into the sum gives

$$\frac12\sum_{k=1}^{20}\bigl(k^2+k\bigr)=\frac12\left(\sum_{k=1}^{20}k^2+\sum_{k=1}^{20}k\right).$$

We now evaluate the two standard summations separately. For the sum of the first $$n$$ natural numbers, the formula is

$$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}.$$

Taking $$n=20$$ we obtain

$$\sum_{k=1}^{20}k=\frac{20\cdot21}{2}=210.$$

For the sum of the squares of the first $$n$$ natural numbers, the formula is

$$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}.$$

Again putting $$n=20$$, we find

$$\sum_{k=1}^{20}k^2=\frac{20\cdot21\cdot41}{6}.$$

Calculate step by step: First $$20\cdot21=420$$. Next $$420\cdot41=17220$$. Finally divide by $$6$$:

$$\frac{17220}{6}=2870.$$

Now substitute the two results back into the big expression:

$$\frac12\left(\sum_{k=1}^{20}k^2+\sum_{k=1}^{20}k\right)=\frac12\left(2870+210\right).$$

Simplify inside the parentheses:

$$2870+210=3080.$$

Multiplying by $$\tfrac12$$ gives

$$\frac12\cdot3080=1540.$$

So, the answer is $$1540$$.