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September 23, 2014

September 23, 2014

Posted by **yumi** on Sunday, March 3, 2013 at 3:38pm.

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

i dont understand when trying to prove tht P(k+1) = 1+2+3+...+k+k(k+1)= 1/2

how do you do this step???

- pre-calculus -
**Steve**, Sunday, March 3, 2013 at 4:00pmThis is called induction. You assume it's true for n=k. Then, if it's true for k, you show that it must also be true for k+1.

Then, show it's true for n=1.

That means it's true for n=2, thus 3, and so on.

So, assume that

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

Now consider

1+2+3+...+k+k+1

Assuming our hypothesis is tur for n=k, that means that

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

This is the crucial step. We already "know" what 1+2+3+...+k is, so we just substitute it in.

Now start rearranging the right hand side:

k(k+1)/2 + k+1 = [k(k+1) + 2(k+1)]/2

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

Wow! This is just our formula, if we substitute in k+1 for k.

If 1+2+3+...+k = f(k)

then 1+2+3+...+k+k+1 = f(k+1)

Now, what if n=1?

1 = 1(1+1)/2 = 1(2)/2 = 1

so our formula is true for n=1.

We have shown that it must also be true for n=2, n=3, ...

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