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anonymous
 one year ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
1^2 + 4^2 + 7^2 + ... + (3n  2)^2 = n(6n^23n1)/2
anonymous
 one year ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. 1^2 + 4^2 + 7^2 + ... + (3n  2)^2 = n(6n^23n1)/2

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0using http://openstudy.com/study#/updates/504b9d7ce4b0985a7a58b494 I understand everything until

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't get how the last form is "precisely what we were supposed to get" and how it proves it true

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3hmmm... first we need the basis then when we have k and then k+1 have to manipulate the left to equal the right ..

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1for n= 1, we have: left side = 1^2 = 1 right side= 1*(631)/2= 1 so our proposition is true for n=1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3ok so we have the basis proven true.. next we need to have n = k

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in general, the mathematical induction principle requests another checking, for example n=2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok I understand why and how to do that

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so, for n=2, we have: left side = 1+ 4^2=1+16=17 right side = 2*(2461)/2= 17

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but what about n = k + 1? how is the simplified form of n = k+1 proving it true???

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3first we need n = k and then we go to n = k+1 . That guy went forward to n+1

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1when we have a proposition which depends on a natural number, say n, namely, we have: P(n), then if P(1) is true and P(n+1) is true when P(n) is true, then we can state, that P(n) is true for all natural numbers, namely: \[\Large \forall n \in \mathbb{N}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[1^2 + 4^2 + 7^2 + ... + (3n  2)^2 = \frac{n(6n^23n1)}{2}\] for n = k \[\[1^2 + 4^2 + 7^2 + ... + (3k  2)^2 = \frac{k(6k^23k1)}{2}\]\] for n = k+1 \[\[1^2 + 4^2 + 7^2 + ... + (3(k+1)  2)^2 = \frac{(k+1)(6(k+1)^23(k+1)1)}{2}\]\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so to simplify this mess a bit ... sorry this is long! \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6(k+1)^23(k+1)1)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6(k+1)^23k4)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6(k^2+2k+1)3k4)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6k^2+12k+6)3k4)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6k^2+12k3k4+6)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6k^2+9k+2)}{2} \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3hmmm then manipulate the left... hold on this is in my book I need to refresh my memory on it.. I remember substituting that 1^2+4^2+7^2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I know that our goal is to have the left = right ... I just have to see one of the examples in my book again.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3a proposition was used... For all n in N dw:1435559494200:dw then the entire 1^2 4^2 7^2 is substituted . we do know that \[(3k+1)^2 = (3k+1)(3k+1) = 9k^2+6k+2 \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I'm going to see what he did on the left.. I know that string of squares get substituted which is what he did. And I remembered that too.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3where the heck did he get ... 6n^23n1 that is my issue right now.. otherwise everything is accounted for..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3he did use a proposition.. so it's the n+1 one in that proposition that's driving me nuts. on top of that .. it's non factor able :/ \[\large =\frac{n(6n^23n1)}{2}+(3n+1)^2 \] that's the part where I'm getting lost. that 6n^23n1 after getting that issue out we need the same denominator so we have 2/2 on the right. and with a bunch of expansion and simplification it will become true.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3the whole string before (3k+1)^2 ends up being replaced by a proposition.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I'm more confused now

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I also need to eat :/ I can't think without carbs.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Take advantage of notation: \(1^2 + 4^2 + 7^2 + ... + (3n  2)^2 = \dfrac{n(6n^23n1)}{2}\). We want to show that \(\sum_1^{n+1}(3n2)^2= \dfrac{(n+1)(6(n+1)^23(n+1)1)}{2}\) given that we know \(\sum_1^n(3n  2)^2=\dfrac{n(6n^23n1)}{2}\). Now \(\sum_1^{n+1}(3n2)^2=(3(n+1)2)^2+\sum_1^n(3n  2)^2=(3n+1)^2+\sum_1^n(3n  2)^2\\= (3n+1)^2+\dfrac{n(6n^23n1)}{2}=\dfrac{(n+1)(6(n+1)^23(n+1)1)}{2}\) and we are done.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0To convince yourself that induction actually works just think of dominoes. If it is the case that when one domino falls, then the next one falls, and if we start at the beginning, then all of the dominos will fall.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3Oh I see what happened. I'd overlooked something We do use substitution. So let \[\large 1^2+4^2+7^2+\dots+(3n2)^2=\frac{n(6n^23n1)}{2} \] then we replace that large chunk \[\large \[1^2 + 4^2 + 7^2 + ... + (3n2)^2+ (3k+1)^2 = \frac{(k+1)(6k^2+9k+2)}{2}\] \[\large \[\frac{n(6n^23n1)}{2}+ (3k+1)^2 = \frac{(k+1)(6k^2+9k+2)}{2}\] then make the left side have the same fraction. But first expand the (3k+1)^2 Recall that \[(3k+1)^2 = (3k+1)(3k+1) = 9k^2+6k+1\] LET n = k so the variables be the same \[\[\large \[\frac{k(6k^23k1)}{2}+ 9k^2+6k+1 \cdot \frac{2}{2} = \frac{(k+1)(6k^2+9k+2)}{2}\]\] \[ \[\[\large \[\frac{(6k^33k^2k)}{2}+\frac{18k^2+12k+2}{2}= \frac{(k+1)(6k^2+9k+2)}{2}\]\]\] now combine like terms on the left \[\large \frac{6k^3+15k^2+11k+2}{2}= \frac{(k+1)(6k^2+9k+2)}{2}\] now I'm going to expand the right to prove that induction holds true. \[(k+1)(6k^2+9k+2) \rightarrow 6k^3+9k^2+2k+6k^2+9k+2 = 6k^3+15k^2+11k+2\] \[\large \frac{6k^3+15k^2+11k+2}{2}= \frac{6k^3+15k^2+11k+2}{2}\] Viola! We have proven that induction holds true for all cases.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3the original problem was sneaky because it had important information. We had to use substitution. It's like saying after you figure out the n =k+1 case replace the entire line of those squares with this \[\large 1^2 + 4^2 + 7^2 + ... + (3n  2)^2 = \frac{n(6n^23n1)}{2}\] and then manipulate the left a bit so left = right And I've also proven to never ever do proofs without meals. I had to bail out because I felt weak and couldn't concentrate anymore, so my mind went somewhere else.
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