Prove that 2 + 4 + 6 + … + 2n = n(n + 1) for all positive integers n. Would appreciate if someone would show me how to solve this one step by step.

- anonymous

- chestercat

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- ganeshie8

what do you know about "mathematical induction" ?

- anonymous

Yes, that is what we are supposed to use. I have been working on induction all weekend. This setup is just different than the others I have been doing. I tried to substitute n for K+1 and I didn't come up with a solution when I did the algebra.

- anonymous

Oh, what do I know*** I'm sorry

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- anonymous

In mathematical induction you prove a base case. Then, you assume an assumption to be true. You put that assumption in your equation. Then you isolate that term from the rest and if it all works then it's true.

- welshfella

Fisrt show that the identity is true for n = 1
plug in n = 1 into right side and see if it comes to 2

- welshfella

1(1 + 1) = ?

- welshfella

= 2 right? so that shows its true for n = 1

- welshfella

r u there?

- welshfella

- anonymous

@welshfella i'm back. thank you for replying. I have the base case. as for the induction step i would assume i'd write that as 2K+1=K+1(K+1+1)

- welshfella

sum of k terms is assumed to be k(k + 1)
now add the (k+1)th term
= k(k + 1) + 2(k + 1)
now you need to manipulate that to get ((k + 1)(k + 1 + 1) which will constitute the proof that if the proposition is true for n = k then its true for n = k+1

- anonymous

okay: so 2k+1 = (k+1)(k+1+1) is the proper set up? That is how I started. There must be something wrong with my algebra then. I will rework the problem on paper and then post what I have here so you can help.

- welshfella

i'm not sure what you mean by that
we have assumed that its true for k terms ( ie k(k+1))
the sum of k + 1 terms would be sum of k terms + (k+1) th term
= k(k + 1) + 2(k + 1)
Now this must be converted to (k+1)(k+1 + 1)

- welshfella

its quite a simple conversion

- anonymous

The Inductive Hypothesis is that n=k
The inductive step is solving it for the next link in the chain, which is k+1. So, when we start our equation n becomes k+1. That's the only conversion I see. Otherwise it's working out 2k+1=k+1(k+1+1)
2k+1 should be on the left side of the equation, no? You have it written together on the right.

- welshfella

hmmm I dont follow that to be honest.
the way i learnt it was as follows( a long time ago !)
1 first prove that the given expression is true for the sum of 1 term.
2. Put n = k and write out the identity in terms of k
We assume this to be true
3. then add the ( k+1) th term
4try and . Manipulate the right side so that it comes to same form as for k with k being replaced by (k+1)

- welshfella

2 + 4 + 6 + … + 2n = n(n + 1)
for n = k we have
2 + 4 + 6 + … + 2k =k(k + 1)
for n = k + 1 we have
2 + 4 + 6 + … + 2k + 2k + 1 = k(k + 1) + 2k + 1
now try to get the right side into the for (k+1)(k + 1 + 1)

- welshfella

* 2k +1 should read 2(k + 1)

- welshfella

2 + 4 + 6 + … + 2k + 2(k + ) = k(k + 1) + 2(k + 1)

- welshfella

- i notice you are writing 2k+1 th term as 2k + 1
- tahts not correct - it is 2(k+1)

- welshfella

so we have
k(k = 1) + 2(k + 1)
- take out the common factor (k+ 1)
we get
(k + 1)(k + 2)
= (k + 1)((k + 1) + 1)
which is the formula for sum of k terms with the k replaced by k + 1

- welshfella

* second line typo its k (k +1) + 2(k +1)

- welshfella

the 2(k+1) is added to both sides of the identity

- welshfella

does this make sense to you?
I must admit I took a while to understand proof by induction.

- anonymous

@welshfella i'm sorry for disappearing yesterday. this proof makes sense to me. i am finally starting to understand. thank you very much for taking the time to do that step by step :)

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