# pre-calculus

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proving that the statements are true for all natural numbers

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

i don't understand when trying to prove that P(k+1) = 1+2+3+...+k+k(k+1)= 1/2

how do you do this step???

• pre-calculus -

This 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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