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anonymous
 one year ago
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
 one year ago
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.

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1what do you know about "mathematical induction" ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, what do I know*** I'm sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In 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
 one year ago
Best ResponseYou've already chosen the best response.2Fisrt show that the identity is true for n = 1 plug in n = 1 into right side and see if it comes to 2

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2= 2 right? so that shows its true for n = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2sum 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
 one year ago
Best ResponseYou've already chosen the best response.0okay: 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
 one year ago
Best ResponseYou've already chosen the best response.2i'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
 one year ago
Best ResponseYou've already chosen the best response.2its quite a simple conversion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.2hmmm 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
 one year ago
Best ResponseYou've already chosen the best response.22 + 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
 one year ago
Best ResponseYou've already chosen the best response.2* 2k +1 should read 2(k + 1)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.22 + 4 + 6 + … + 2k + 2(k + ) = k(k + 1) + 2(k + 1)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2 i notice you are writing 2k+1 th term as 2k + 1  tahts not correct  it is 2(k+1)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2so 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
 one year ago
Best ResponseYou've already chosen the best response.2* second line typo its k (k +1) + 2(k +1)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2the 2(k+1) is added to both sides of the identity

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2does this make sense to you? I must admit I took a while to understand proof by induction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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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