Where is wrong again?

Algebra Level pending

Watch the flow and see where is the wrong.

\(------\)

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?

\(-----\)

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

×

Problem Loading...

Note Loading...

Set Loading...