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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?

A6SE
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    See https://en.wikipedia.org/wiki/Telescoping_series – lab bhattacharjee Oct 21 '15 at 12:56
  • I just proved the second example by subtracting cubes of k+1 and k and then passing them through sum from 1 to n, and got the correct solution. But, I still don't understand this method. – A6SE Oct 21 '15 at 13:32
  • Look Alex's solution in this thread for my 3rd example, he used the same method as my teacher and I don't know how they came up with the method, I don't want to learn it by heart: http://math.stackexchange.com/questions/973242/sum-of-cubes-proof – A6SE Oct 21 '15 at 14:31
  • I suggest that you study the chapter 3 of the Polya's "Mathematical Discovery" and as he said "Do you wish to know what is behind the trick? Try to apply that trick yourself and then you may find out." – Woria Oct 07 '17 at 19:22

2 Answers2

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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

$2S = n(n+1) \Rightarrow S = \frac{n(n+1)}{2}$

IrbidMath
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Define the sequence $\{S_k(n)\}_{k\geq0}$ as follows: $S_k(1)=1;~S_0(n)=n$, and \begin{equation}\label{e1} S_{k+1}(n)=(k+1)\int_0^n S_k(x)dx+n\Bigl(1-\int_0^1 S_k(x)dx\Bigr),~k=0,1,\ldots \end{equation} Now you can derive the following formulas:

  • $S_0(n)=n$
  • $S_1(n)=\frac{n(n+1)}{2}$
  • $S_2(n)=\frac{n(n+1)(2n+1)}{6}$
  • $S_3(n)=\frac{n^2(n+1)^2}{4}$
  • $\ldots$
Singh
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