# Where is wrong again?

Algebra Level pending

Watch the flow and see where is the wrong.

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We need to check that $$1+2+3+4+5+6+...+n = \cfrac{n(n+1)}{2}$$

1) Let S = $$\cfrac{n(n+1)}{2}$$

2) $$1+2+3+4+5+6+...+n = S$$

3) $$1+(1+1)+(1+2)+(1+3)+...+(1+n-1) = S$$

4) $$1+1+1+...+1+1+2+3+4+5+...+n-1 = S$$ where 1+1+1+1+...+1 = n 1's

5) $$n+1+2+3+4+5+6+n-1 = S$$

6) $$1+2+3+4+...+n-1 = S-n$$

7) $$(1+2+3+4+...+n)-1 = S-n$$

8) $$S-1 = S-n$$

9) $$n = 1$$

10) $$\therefore 1+2+3+4+5+...+n = S = \cfrac{n(n+1)}{2} = \cfrac{1(1+1)}{2} = 1?!$$

Where is the start of the wrong?

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Note: Just put the number that corresponds to the statement that is wrong.

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