Induction q (1 Viewer)
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ExtremelyBoredUser
Bored Uni Student
Note: The induction procedure is different for every person so I'm just going to focus on the solution itself, ask your teacher for how they'd like it when you start year 11.
[Base Case]
When there is 3 sides, you will form a triangle and this has no diagonals.
Hence, n = 3 is a true statement.
[Induction Hypothesis]
Suppose k is an integer and true for n >= 3.
n = k;
Polygon with k sides will have
[Proving stage]
n = k + 1;
Claim/Proof: We have to prove a k+1 side polygon has
From induction hypothesis, a polygon with k sides will have
Here are some really shoddy diagrams to give an idea of what I'm talking about. The red diagonal (please ignore the fact the diagonal's not straight) is the diagonal thats unaccounted for from going from n = 5 to n = 6 (you can consider this as k = 5) and the yellow vertex is the new vertex. You can see k - 2 ( 6 - 2 = 4) diagonals are formed solely from the diagonals however the edges are not considered.
n = 5 to n = 6.
This the logic we're using for n = k + 1.
...
Hence proven by principles of mathematical induction for n>=3 yadda yadda this part is whatever your teacher wants etc.
For these type of qs, I would just try to draw diagrams for the cases and if you still don't get the pattern, draw for more cases and it'll become apparent (as above for n = 5 to n = 6) and then proceed. I think there's similar type qs like proving by induction that n lines divding the plane into certain amount regions or smthn like that.
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ohh makes sense now, I also had another question to do with induction with inequalities, I saw this step in one worked example somewhere was the 1+1/2k able to be discarded as since k>1 removing the expression just makes the rhs of a lesser value?Note: The induction procedure is different for every person so I'm just going to focus on the solution itself, ask your teacher for how they'd like it when you start year 11.
[Base Case]
When there is 3 sides, you will form a triangle and this has no diagonals.
Hence, n = 3 is a true statement.
[Induction Hypothesis]
Suppose k is an integer and true for n >= 3.
n = k;
Polygon with k sides will havediagonals
[Proving stage]
n = k + 1;
Claim/Proof: We have to prove a k+1 side polygon hasdiagonals
From induction hypothesis, a polygon with k sides will have
Here are some really shoddy diagrams to give an idea of what I'm talking about. The red diagonal (please ignore the fact the diagonal's not straight) is the diagonal thats unaccounted for from going from n = 5 to n = 6 (you can consider this as k = 5) and the yellow vertex is the new vertex. You can see k - 2 ( 6 - 2 = 4) diagonals are formed solely from the diagonals however the edges are not considered.
View attachment 34655 View attachment 34654
n = 5 to n = 6.
This the logic we're using for n = k + 1.
...
Hence proven by principles of mathematical induction for n>=3 yadda yadda this part is whatever your teacher wants etc.
For these type of qs, I would just try to draw diagrams for the cases and if you still don't get the pattern, draw for more cases and it'll become apparent (as above for n = 5 to n = 6) and then proceed. I think there's similar type qs like proving by induction that n lines divding the plane into certain amount regions or smthn like that.
ExtremelyBoredUser
Bored Uni Student
Yeah pretty much, the value of
(you can consider it like a constant which is > 1 and so the magnitude of 1/2k+1 could not be less than 1/2k+1 for k > 1 as stated in the q)
so therefore the RHS in line 1 would be greater than
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ohhhh okay thank youYou were really close.
(a-b)\geq{0}\\\\a^n\cdot{a}-a^nb-b^na+b^n\cdot{b}\geq{0}\\\\a^{n+1}+b^{n+1}\geq{a^nb+b^na})