Let f(n)=1+2+3+...+n.
Now, we have f(n)−f(n−1)=n, and we will prove f(n)+f(n−1)=n2 by induction.
Since the proposition is true for n=1, let us assume it is true for n=k. Now for n=k+1, we have: f(k)+f(k+1)=[f(k)+f(k−1)]+2k+1=k2+2k+1=(k+1)2. So, we conclude that f(n)+f(n−1)=n2 for n belongs to N. Now, we have a system of simultaneous equations: f(n)+f(n−1)=n2 f(n)−f(n−1)=n solving them, we have 2f(n)=n2+n which yields:
f(n)=n(n+1)/2
Easy Math Editor
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Top NewestI guess you meant f(n)-f(n-1)=n (not 2) at two instances.
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thank you very much!!!
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Cool
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thank you very much
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How is this proof?
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Well I guess that you used the mathematical induction in this problem, right? It's a kind of simple method to prove such these kind of mathematical statements. I like it.
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Thx!!!
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Cool proof! @Mohammed Imran
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thank you!
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Oh nice, where did you see this?
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this is the original proof of the theorem by gauss
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Then why did you type it as yours?🤔
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@Brilliant Mathematics, @joan davis has wrote 5 posts that don't relate to mathematics.
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