1+2+...+n+(n-1)+...+1 = n^2 증명

Theorem

∀n∈N:1+2+⋯+n+(n−1)+⋯+1=n 2

Illustration

Direct Proof 1

							1+2+⋯+(n−1)+n+(n−1)+⋯+1
					=		1+2+⋯+(n−1)+(n−1)+⋯+1+n
					=		2(1+2+⋯+(n−1))+n
					=		2((n−1)n 2 )+n			Closed Form for Triangular Numbers
					=		(n−1)n+n
					=		n 2 −n+n
					=		n 2			as desired.

■

Direct Proof 2

							1+2+⋯+(n−1)+n+(n−1)+⋯+1
					=		(1+2+⋯+(n−1))+(1+2+⋯+(n−1)+n)
					=		(n−1)n 2 +n(n+1) 2			Closed Form for Triangular Numbers
					=		n 2 −n+n 2 +n 2
					=		n 2			as desired.

■

Proof by Induction

Proof by induction:

Base case

n=1 holds trivially.
Just to make sure, we try n=2 :

1+2+1=4

Likewise n 2 =2 2 =4 .
So shown for base case.

Induction Hypothesis

This is our induction hypothesis:

1+2+⋯+k+(k−1)+⋯+1=k 2

Now we need to show true for n=k+1 :

1+2+⋯+(k+1)+k+(k−1)+⋯+1=(k+1) 2

Induction Step

This is our induction step:

							1+2+⋯+(k+1)+k+(k−1)+⋯+1
					=		(1+2+⋯+k+(k−1)+⋯+1)+k+(k+1)
					=		k 2 +k+(k+1)			from induction hypothesis
					=		k 2 +2k+1
					=		(k+1) 2

The result follows by induction.
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