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^2-3n-1)/2

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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^2-3n-1)/2

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I don't get how the last form is "precisely what we were supposed to get" and how it proves it true
hmmm... first we need the basis then when we have k and then k+1 have to manipulate the left to equal the right ..

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for n= 1, we have: left side = 1^2 = 1 right side= 1*(6-3-1)/2= 1 so our proposition is true for n=1
ok so we have the basis proven true.. next we need to have n = k
in general, the mathematical induction principle requests another checking, for example n=2
Ok I understand why and how to do that
so, for n=2, we have: left side = 1+ 4^2=1+16=17 right side = 2*(24-6-1)/2= 17
but what about n = k + 1? how is the simplified form of n = k+1 proving it true???
first we need n = k and then we go to n = k+1 . That guy went forward to n+1
when 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: \[\Large \forall n \in \mathbb{N}\]
\[1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{n(6n^2-3n-1)}{2}\] for n = k \[\[1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = \frac{k(6k^2-3k-1)}{2}\]\] for n = k+1 \[\[1^2 + 4^2 + 7^2 + ... + (3(k+1) - 2)^2 = \frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]\]
so to simplify this mess a bit ... sorry this is long! \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6(k+1)^2-3k-4)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6(k^2+2k+1)-3k-4)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6k^2+12k+6)-3k-4)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6k^2+12k-3k-4+6)}{2} \] \[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6k^2+9k+2)}{2} \]
hmmm 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
I know that our goal is to have the left = right ... I just have to see one of the examples in my book again.
a 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 \[(3k+1)^2 = (3k+1)(3k+1) = 9k^2+6k+2 \]
I'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.
Ok
where the heck did he get ... 6n^2-3n-1 that is my issue right now.. otherwise everything is accounted for..
he 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 :/ \[\large =\frac{n(6n^2-3n-1)}{2}+(3n+1)^2 \] that's the part where I'm getting lost. that 6n^2-3n-1 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.
the whole string before (3k+1)^2 ends up being replaced by a proposition.
I think I'm more confused now
I also need to eat :/ I can't think without carbs.
Take advantage of notation: \(1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \dfrac{n(6n^2-3n-1)}{2}\). We want to show that \(\sum_1^{n+1}(3n-2)^2= \dfrac{(n+1)(6(n+1)^2-3(n+1)-1)}{2}\) given that we know \(\sum_1^n(3n - 2)^2=\dfrac{n(6n^2-3n-1)}{2}\). Now \(\sum_1^{n+1}(3n-2)^2=(3(n+1)-2)^2+\sum_1^n(3n - 2)^2=(3n+1)^2+\sum_1^n(3n - 2)^2\\= (3n+1)^2+\dfrac{n(6n^2-3n-1)}{2}=\dfrac{(n+1)(6(n+1)^2-3(n+1)-1)}{2}\) and we are done.
To 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.
thank you
Oh I see what happened. I'd overlooked something We do use substitution. So let \[\large 1^2+4^2+7^2+\dots+(3n-2)^2=\frac{n(6n^2-3n-1)}{2} \] then we replace that large chunk \[\large \[1^2 + 4^2 + 7^2 + ... + (3n-2)^2+ (3k+1)^2 = \frac{(k+1)(6k^2+9k+2)}{2}\] \[\large \[\frac{n(6n^2-3n-1)}{2}+ (3k+1)^2 = \frac{(k+1)(6k^2+9k+2)}{2}\] 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 \[\frac{k(6k^2-3k-1)}{2}+ 9k^2+6k+1 \cdot \frac{2}{2} = \frac{(k+1)(6k^2+9k+2)}{2}\]\] \[ \[\[\large \[\frac{(6k^3-3k^2-k)}{2}+\frac{18k^2+12k+2}{2}= \frac{(k+1)(6k^2+9k+2)}{2}\]\]\] 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.
the 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 \[\large 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{n(6n^2-3n-1)}{2}\] 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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