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???
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, ...
To prove the statement for all natural numbers, we use mathematical induction.
First, we prove the base case, which is when n = 1.
When n = 1, the left-hand side (LHS) is 1, and the right-hand side (RHS) is 1(1+1)/2 = 1. Since LHS = RHS, the statement holds true for n = 1.
Now, let's assume that the statement holds true for some value k, which means that 1 + 2 + 3 + ... + k = k(k + 1)/2.
Next, we need to prove that the statement also holds true for k + 1. So, we want to show that 1 + 2 + 3 + ... + k + (k + 1) = (k + 1)(k + 1 + 1)/2.
To do this, we add (k + 1) to both sides of the equation for the assumption:
1 + 2 + 3 + ... + k + (k + 1) = k(k + 1)/2 + (k + 1).
Now, we can simplify the right-hand side:
k(k + 1) + 2(k + 1) = (k + 1)(k + 2).
Notice that the right-hand side of the assumed equation is the same as the right-hand side of the statement we're trying to prove for k + 1.
Therefore, by assuming the statement holds true for k and then showing that it holds true for k + 1, we conclude that the original statement is true for all natural numbers by mathematical induction.