I'm not sure how to prove some of the identities without using induction, for example: $$1+2+3+...+n=\frac{n(n+1)}{2}$$ $$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$ $$1^3+2^3+...+n^3=(\frac{n(n-1)}{2})^2$$ What my teacher suggested and did for the first example is, take two successive members and sum their squares, then using some transformations get $\frac{n(n+1)}{2}$, here's what the teacher did: $$(k+1)^2-k^2=2k+1$$ we sum $k$-s from $1$ to $n$: $$\sum_{k=1}^{n}((k+1)^2-k^2)=2\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$$ $$2^2-1^2+3^2-2^2+4^2-3^2+...+(n+1)^2-n^2=2\sum_{k=1}^{n}k+n$$ $$(n+1)^2-1=2\sum_{k=1}^{n}k+n$$ $$n^2+2n-n=2\sum_{k=1}^{n}k$$ $$2\sum_{k=1}^{n}k=n^2+n$$ $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$
What is the method my teacher used here? The teacher also suggested that, for example, if we have the sum of squares of successive integers (like in the second example), we should take two successive numbers and sum their cubes, or if we have the sum of cubes, then we take two successive members of the sum and sum their 4th degrees. Is there a name for this method of solving so I could google it and examine it a bit more?
For the first one there is a nice way to find it is sum let
$S = 1 + 2 + 3 \cdots + n $ then
$S = n + (n-1) + (n-2) + \cdots + 2 + 1 $ add them by terms (first term with the first second with the second we get
$2S = (n+1) + (n+1) + (n+1) \cdots + (n+1) $ but we have $n$ of $(n+1)$ so
$2S = n(n+1) \Rightarrow S = \frac{n(n+1)}{2}$