Find a general formula for the following sums:

1. $\sum_{k=1}^{n} k^{3}$
2. $\sum_{k=1}^{n} 2 k-1$
3. $28+31+34+\cdots+253$
1. Let's look at the first values: $1,9,36,100, \ldots$.

We notice that these are the squares of the numbers: $1,3,6,10, \ldots$

We recognize the list of triangular numbers, so we can conjecture that:

$$
\sum_{k=1}^{n} k^{3}=\left(\sum_{k=1}^{n} k\right)^{2}
$$

I leave it to the reader to prove this result by induction.

2. Idea: the sum is linear, that is, if $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ and $\lambda$ (lowercase lambda) are real numbers, then:

$$
\sum_{k=1}^{n}\left(a_{k}+\lambda b_{k}\right)=\sum_{k=1}^{n} a_{k}+\lambda \sum_{k=1}^{n} b_{k}
$$

Therefore:

$$
\begin{aligned}
& \sum_{k=1}^{n} 2 k-1=2 \sum_{k=1}^{n} k-\sum_{k=1}^{n} 1 \\
& =2 \frac{n(n+1)}{2}-n \\
& =n^{2}
\end{aligned}
$$

3. We notice that $28+31+\cdots+253=\sum_{k=1}^{76} 3 k+25$ and we use the linearity of the sum. Or, we notice that:

$$
\begin{aligned}
28+31+\cdots+253 & =\frac{1}{2}((28+253)+(31+250)+\cdots+(253+28)) \\
& =\frac{1}{2}(76 \times 281) \\
& =10678
\end{aligned}
$$

Because there are 76 terms $\left(\frac{253-28}{3}+1=76\right.$, don't forget the " $\left.+1 "!\right)$.