Alternate Proof to Induction for Integer Sum Another way to prove the expression for the sum of all positive integers up to and including n
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- in the last video we proved that
- the sum of all of the positive integers, up to and including n
- can be expressed as n(n+1)/2
- and we proved that by induction
- what I want to do in this video is show you that there is actually a simpler proof of that
- but it's not by induction
- so it wouldn't be included in that video, but I'll show you that it exists.
- just so that you know that induction isn't the only way to prove it
- so we defined that fuction s(n)
- as the sum of all of the positive integers up to n including n
- so this is equal to by definition
- 1+ 2+ 3 all the way to + (n-1) + n
- so it's all the integers up to and including n
- this is how we're defining it
- well we can re-write it again
- we can say that the sum S(n); we can write the same thing
- but we can write it differently order
- we can say that this is the same thing as n +( n -1 )+( n - 2 )
- plus all the way down to + 2 + 1
- well what does this do for us
- we can actually add these two rows
- if we add S(n) + S(n) we're going to get 2 times the sum
- so we're just adding on the left
- and then we can add on the right
- so we're really just doing this sum twice
- and what's interesting is how we're going to add it
- we're going to add this term to this term; this term to this term
- we're just trying to add these two things
- we can pick any way we want to add them
- (1 + n) is going to be( n + 1)
- and then we're going to add 2 +( n - 1)
- so what's that?
- Let me write that over here
- 2 +( n - 1)
- is the same things as 2 + n - 1,
- which is the same thing as n + 1
- so this is also going to be n+1
- and this term over here, 3 + n - 2
- or n - 2 +3
- that again is going to be n +1
- and you're going to do that for every term until you get over here
- n -1 +2; that's also going to be n+1
- and then finally you have n+1 right over here
- so what's this whole sum going to be
- well how many of these n + 1's do we have?
- well we have n of them
- for every term in each of these sums
- So there's 1, 2,3 count all of the way to n
- So you have n of these n+1's
- so if you add something to its self n times
- So you really have, this is exactly equivalent to
- n times (n+1)
- so two times the sum of all of the positive integers
- up to an including in is going to be equal to n(n+1)
- so if you divide both sides by 2 we get an expression for the sum
- so the sum of the integers is going to be equal to
- n(n+1) over 2
- so here's a proof where we didn't have to use induction
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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