Hi there,
Let P(n) denote a proposition depending on n. For instance:
P(n): 2+4+6+....+2n = n(n+1)
for n=1, 2, ...
To prove that P(n) is true for every integer n, we do the following
steps:
Verify P(1) (basis step)
Prove that is P(k) is true, then so is P(k+1) (induction step)
Therefore, one proves P(1), then P(2), then P(3), etc... until
infinity.
For the question you've provided, the proof is as follows:
.: Basis step (show that P(1) is true)
P(1): 2=1(1+1)
=2
P(1) is verified.
.: Induction step. (show that if P(k) is true, P(k+1) is also true)
->Assume<- P(k) is true. This means that
2+4+6+...+2k = k(k+1)
Show that P(k+1) is true, that is
2+4+6+...+2k+2(k+1)=(k+1)(k+2)
To do this, start with the left side of P(k+1) and then modify to
equivalent statements until we get the right side of P(k+1)
2+4+6+...+ 2k + 2(k+1)
= (2+4+6+...+2k)+2(k+1)
= k(k+1) + 2(k+1) <- the left side of P(k) was replaced with the
right side (remember the assumption at the start of the induction
step)
= (k+1)(k+2) <- factored out the common term (k+1)
Thus the induction step is complete.
Therefore, by the principle of mathematical induction, it has been
shown that 2 + 4 + 6 + ... + 2n = n(n+1) whenever n is a positive
integer.
If you need clarification on any part of the answer, please request
clarification.
Best regards,
Tox-ga |
