How do you prove something by mathematical induction?

Question

I know that you have to check with 1, first, then plug k in, and finally k+1. If something is true, is your k+1 equation supposed to be equal to the k equation?

Answer 1

What you try to do is show that IF it's true for some value k, then it has to be true for k+1. Then simply show manually that it's true for k=1. Because that now implies it's true for 2, which implies it's true for 3, which implies it's true for 4, etc.

For example, if we wanted to prove that the sum of all integers from 1 to n is n(n+1)/2, we'll first assume that this is true:
1 + 2 + 3 + ... + k = k(k+1)/2

To show that it's true for k+1, add k+1 to both sides and simplify:
1 + 2 + ... + k + (k+1) = k(k+1)/2 + (k+1)
1 + 2 + ... + k + (k+1) = k(k+1)/2 + 2(k+1)/2
1 + 2 + ... + k + (k+1) = [ k(k+1) + 2(k+1) ] /2
1 + 2 + ... + k + (k+1) = [ (k+1)(k + 2) ] /2
1 + 2 + ... + k + (k+1) = [ (k+1)(k+1 + 1) ] /2

Notice the formula in the right is in the form of n(n+1)/2. So we proved that if it's true for k, then it's true for k+1. Finally, notice that 1 = 1*(1+1)/2, which means the formula holds for n=1. This means it holds for n=2, which means it holds for n=3, etc.

Answer 2

No, your k+1 equation just has to be true. It will usually use your k equation (which you just assume to be true).

Answer 3

No, but your k+1 equation would be true -assuming- that the k equation is true.
For more info see..
http://en.wikipedia.org/wiki/Mathematical_induction
http://www.cut-the-knot.org/induction.shtml
http://www.earlham.edu/~peters/courses/logsys/math-ind.htm