A New Proof Of 1+2+3+...+n = n(n+1)/2n(n+1)/2

Let f(n)=1+2+3+...+nf(n)=1+2+3+...+n.

Now, we have f(n)f(n1)=nf(n)-f(n-1)=n, and we will prove f(n)+f(n1)=n2f(n)+f(n-1)=n^2 by induction.

Since the proposition is true for n=1n=1, let us assume it is true for n=kn=k. Now for n=k+1n=k+1, we have: f(k)+f(k+1)=[f(k)+f(k1)]+2k+1=k2+2k+1=(k+1)2f(k)+f(k+1)=[f(k)+f(k-1)]+2k+1=k^2+2k+1=(k+1)^2. So, we conclude that f(n)+f(n1)=n2f(n)+f(n-1)=n^2 for n belongs to N. Now, we have a system of simultaneous equations: f(n)+f(n1)=n2f(n)+f(n-1)=n^2 f(n)f(n1)=nf(n)-f(n-1)=n solving them, we have 2f(n)=n2+n2f(n)=n^2+n which yields: f(n)=n(n+1)/2f(n)= \boxed {n(n+1)/2}

Note by Mohammed Imran
6 months, 2 weeks ago

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I guess you meant f(n)-f(n-1)=n (not 2) at two instances.

K T - 6 months, 2 weeks ago

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thank you very much!!!

Mohammed Imran - 6 months, 2 weeks ago

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Cool

Nabeel Akhter - 5 months, 4 weeks ago

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thank you very much

Mohammed Imran - 5 months, 4 weeks ago

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How is this proof?

Mohammed Imran - 5 months, 4 weeks ago

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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.

Anh Khoa Nguyễn Ngọc - 5 months, 2 weeks ago

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Thx!!!

Mohammed Imran - 5 months, 2 weeks ago

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Cool proof! @Mohammed Imran

A Former Brilliant Member - 4 months, 1 week ago

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Oh nice, where did you see this?

Nitin Kumar - 6 months, 1 week ago

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this is the original proof of the theorem by gauss

Mohammed Imran - 6 months, 1 week ago

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Then why did you type it as yours?🤔

Nitin Kumar - 6 months, 1 week ago

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@Nitin Kumar I said chew seong's proof is not original

Mohammed Imran - 6 months, 1 week ago

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@Mohammed Imran Ok

Nitin Kumar - 6 months, 1 week ago

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@Brilliant Mathematics, @joan davis has wrote 5 posts that don't relate to mathematics.

A Former Brilliant Member - 3 months, 2 weeks ago

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Thank you for notifying us again. We will take action on this shortly.

Brilliant Mathematics Staff - 3 months, 2 weeks ago

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