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Hey everyone,
Ive been trying to get my head around induction but I am having a hard time working around it. Im stuck on the following problem:
Prove that the sum of the first n odd integers is equal to n^2.
Anyone have any insight as to how to set this up using mathetmatical induction?
Last edited by MarkusD (2007-09-15 08:21:22)
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Last edited by Daniel123 (2007-09-15 08:14:11)
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a)1=1²
-the proposition regarding n=1 case is true.
b)If the proposition for k - k can be any from 1, 2, 3...- is true, that is to say
1+3+...+(2k-1)=k²,
THEN adding one 2k+1 on both sides also stands:
1+3+...+(2k-1)+(2k+1)=k²+2k+1
=> 1+3+...+(2k+1)= (k+1)²
meaning the proposition for k+1 case is true.
Conclusion: a) says the propostion stands for n=1,
b) says the propostion stands for n=k+1 on the assumption that it stands for n=k.
Thus, starting from n=1, using b) recursively,
n=1 true => n=2 true
n=2 true => n=3 true
n=3 true => n=4 true
.......
we can fairly conclude that for any given n, the proposition that
the sum of the first n odd integers is equal to n^2
always stands.
Or, alternatively, it always stands no matter how many terms there are in the sum.
(This is just the demonstration and you don't need to write so long for your assignment )
X'(y-Xβ)=0
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Note to Daniel: That's a flawless proof and it would normally be correct, but the question specifies that it wants a proof by induction, so it needs to be George's way.
Why did the vector cross the road?
It wanted to be normal.
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Wow, quick response. Its interesting to see Daniel's method too. Thanks a lot guys!
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Haha oops!
I didn't read the whole post.. I only looked at the question!
Sorry!
I'm happy my method is still right though, I've only just learned series
Last edited by Daniel123 (2007-09-15 10:16:32)
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