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Is this proof correct, as I feel unsure about whether or not I did that correct because the book did it differently, I wouldn't know however why my proof should be wrong.

Could you help me out?

The following statement is to be proven by induction. $$\sum^n_{i=1} = \frac{n^2+n}{2}$$ Base case $n=1$ $$1 = \frac{1+1}{2} \checkmark $$ Induction Step $n\rightarrow n+1$ $$\sum^{n+1}_{i=1}=\sum^n_{i=1}+(n+1)\\ \iff \frac{n^2+n}{2}+(n+1) \\ \iff \frac{n^2+n}{2}+\frac{2(n+1)}{2} \\ \iff \frac{n^2+n+2n+2}{2}\\ \iff \frac{n^2+3n+2}{2} \\ \iff \frac{(n+1)^2+(n+1)}{2} $$

thebilly
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  • Your proof is correct. – Jochen Dec 17 '18 at 07:47
  • Let try to improve you question using MathJax. – user Dec 17 '18 at 07:50
  • @thebilly have you seen Gauss' solution to this problem, it is quite elegant. – Mustafa Said Dec 17 '18 at 23:34
  • @gimusi Yes I could do that, but that is a lot of work... I'd spend more time writing this into mathjax than solving my problems, which i guess is rather bad time management in regards to exams. is it not ok to post quick questions in pictures? – thebilly Dec 18 '18 at 20:01
  • @thebilly Your question is good, you have shown your work and indeed you have received also a good number of upvotes. Anyway your question should be edited using Mathjax according to the guidelines How to ask a good question.. That's important to create and share a collection of questions and answers which can be used as source and reference also by users in the future. – user Dec 18 '18 at 21:11
  • @thebilly Moreover it could be also useful for your career improve your skills in mathematical typing. – user Dec 18 '18 at 21:13
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    @gimusi i see. I'll edit it. – thebilly Dec 18 '18 at 21:39
  • @thebilly That's nice! Do not hesitate to ask if you need some help. Note that you can past and copy the text you need from any source by copy and paste. – user Dec 18 '18 at 21:41
  • @gimusi thank you. I'd have one question. Do I use the $ \iff $ as the sign to show following transformations?(I am unsure because it says this is the logically equivalent sign and I guess I want to show algebraic equaivalence... – thebilly Dec 18 '18 at 21:48

1 Answers1

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Yes it is correct indeed, also according to the usual way to write that foundamental identity, that is

$$\sum^n_{i=1} i= \frac{n(n+1)}{2}$$

at the end we obtain

$$\sum^{n+1}_{i=1} i= \frac{n^2+3n+2}{2}= \frac{(n+1)(n+2)}{2}$$

Refer also to the related

user
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