Hey guys, I'm trying to do some revision work on 3U, and I've come across a question I have no idea how to finish. Any help (especially with the finishing, getting claim(k+1) to work) would be massively appreciated. Here she is:-
Prove, using mathematical induction, that the total number of diagnonals in an n-sided polygon is given by
n(n-3)
2
Sorry for the poor formating, I couldn't get latex to work.
Attempted Proof
i) Smallest meaningful n = 4
A quadrilateral (n = 4) obviously has 2 diags; 4(4-3)/2 = 2
Therefore true for n = 4
ii) Assume true for n = k (for k >= 4)
i.e. for a k-sided polygon, no of diagonals = k(k-3)/2
iii) To construct a (k+1)-sided polygon, replace any one side of a k-sided
one with 2 sides; now the replaced side is now 1 diagonal and from
the new vertex, we have now (k-2) additional diags to the pre-existing
vertices, That is, in adding one extra side we've added (k-2) + 1 diags.
Therefore for this (k+1)-sided polygon we have k(k-3)/2 [
from assumption]
+ (k - 1) diagonals. Now k(k-3)/2 + k-1 = (k^2 - k -2)/2 =
(
k+1)(
k+1 - 3)/2 diagonals. i.e. the formula holds true for a (k+1)-sided
polygon . . since the formula is the same as given, with 'k' replaced by
'k+1'
1v) Therefore by the Principle of Mathematical Induction, the formula is true
for all n = 4, 5, 6, 7 . . . . No need to say, as is often taught here,
things like "
if true for n = 4, it's true for n = 5, if true for n = 5, then
true for n = 6, . . . and therefore blah blah blah "
PS: Aren't you a little early to be already doing revision for 3U Maths Induction since your HSC is in 2010 ?
