On Tue, 06 May 2008 13:03:28 EDT, Lasonia
<ljkm2006@[EMAIL PROTECTED]
> wrote in
<news:7149825.1210093438837.JavaMail.jakarta@[EMAIL PROTECTED]
>
in alt.math.undergrad:
> I have to create a proof by mathematical induction that
> demonstrates that the sum of the first n even numbers is
> equal to n(n +1).
> Can anyone help a.s.a.p.
You have to understand the basic idea behind mathematical
induction. For the form that you're probably working with
right now, it's the following.
First, you have some statement P(n) about positive integers:
for any positive integer n, P(n) is 'the number n has
such-and-so property'. In your case P(n) is 'the sum of the
first n positive even integers is n(n + 1)'. Thus, P(1) is
the statement '2 = 1 * (1 + 1)', which is certainly true.
P(2) is the statement '2 + 4 = 2 * (2 + 1)', which is also
easily seen to be true.
You want to prove that this P(n) is true for all positive
integers n. To do this by mathematical induction, you must
do two things. First, you must verify that P(1) is true; I
did that in the previous paragraph. Then you must do what
is usually the harder part: you must show that for any
positive integer n, IF P(n) is true, THEN P(n+1) is also
true. You are NOT proving outright that P(n+1) is true; you
are proving only that it must be true IF P(n) is true.
Typically one does this by assuming that P(n) is true for
some definite but unspecified value of n and somehow
deducing from this that P(n+1) must also be true. In your
case this would mean letting n be an arbitrary positive
integer, assuming that P(n) is true, i.e., that the sum of
the first n positive even integers is n(n + 1), and somehow
using this to deduce that P(n+1) is true, i.e., that the sum
of the first n + 1 positive even integers is ... what?
What, exactly, does P(n+1) say for your problem?
This is the first place where many beginners go astray.
P(n+1) is the statement that results when you replace every
instance of n in the statement P(n) by n + 1.
P(n): the sum of the first n positive even integers is
n(n + 1)
P(n+1): the sum of the first n + 1 positive even
integers is (n+1)[(n+1) + 1], i.e., (n + 1)(n + 2)
In short, you want to show that if n is a positive integer
for which it's true that the sum of the first n positive
even integers is n(n + 1), then it will automatically also
be true that the sum of the first n + 1 positive even
integers is (n + 1)(n + 2). The first n positive even
integers are 2, 4, 6, ..., 2n, so in symbols this is:
IF
2 + 4 + 6 + ... + 2n = n(n + 1), [*]
THEN
2 + 4 + 6 + ... + 2n + 2(n+1) = (n + 1)(n + 2). [**]
Now 2 + 4 + 6 + ... + 2n + 2(n+1) can be grouped as
[2 + 4 + 6 + ... + 2n] + 2(n+1),
and you can use the induction hypothesis -- that's the
assumption that I labelled [*] -- to simplify the quantity
in square brackets. With a little more algebra you should
end up with a calculation that looks something like this:
2 + 4 + 6 + ... + 2n + 2(n+1) =
[2 + 4 + 6 + ... + 2n] + 2(n+1) =
<some expression involving n> =
(n + 1)(n + 2).
The first equals sign is justified by basic algebra, the
second by using the hypothesis that [*] is true, and the
third by basic algebra. Equating the first and last
expressions in this chain of equalities gives you the
desired conclusion, [**].
When you've done this, you really will have created a proof
that if P(n) (i.e., [*]) is true, then so is P(n+1) (i.e.,
[**]).
Brian
|
