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n(n+1)/2

You can see this from the following:

Let x=1+2+3+...+n

This is the same as x=n+(n-1)+...+1

x=1+2+3+...+n

x=n+(n-1)+...+1

If you add the corresponding terms on the right-hand side of the two equations together, they each equal n+1 (e.g., 1+n=n+1, 2+n-1=n+1, ..., n+1=n+1). There are n such terms. So adding the each of the left-hand sides and right-hand sides of the two equations, we get:

x+x=(n+1)+(n+1)+...+(n+1) [with n (n+1) terms on the right-hand side

2x=n*(n+1)

x=n*(n+1)/2

A more formal proof by induction is also possible:

(1) The formula works for n=1 because 1=1*2/2.

(2) Assume that it works for an integer k.

(3) Now show that given the assumption that it works for k, it must also work for k+1.

By assmuption, 1+2+3+...+k=k(k+1)/2. Adding k+1 to each side, we get:

1+2+3+...+k+(k=1)=k(k+1)/2+(k+1)=k(k+1)/2+2(k+1)/2=(k(k+1)+2(k+1))/2=((k+2)(k+1))/2=(((k+1)+1)(k+1))/2=((k+1)((k+1)+1)/2

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โˆ™ 2009-12-30 19:12:02
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