Use mathematical induction to prove the statement is true for all positive integers n. 8 + 16 + 24 + ... + 8n = 4n(n + 1)

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Use mathematical induction to prove the statement is true for all positive integers n. 8 + 16 + 24 + ... + 8n = 4n(n + 1)

Mathematics
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Is the statement true for \(n=1\) ?
yes

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Other answers:

how, can u explain
4(1)(1+1) 4(2) =8
that means the expression \(4n(n+1)\) correctly gives the sum on left hand side for \(n=1\)
what if you have 2 terms on left hand side : \(8+16\) is the statement still true ?
wait... you lost me. i have to plug in both now?
you don't need to, one base case is sufficient here, do you know how induction works ?
i'm a little rusty on induction.
basically your goal is to prove that the given statement is true for all positive integers : \(n=1,2,3,\ldots \)
i understand that. but I get confused after I've already proven 1
i don't know where to go from there
suppose, you "know" that the given statement is true for \(n=k+1\), whenever it is true for \(n=k\)
is that sufficient to claim that the given statement is true for all integers : \(n=1,2,3,\ldots\) ?
so then i have to plug in k+1
forget about the process, il surely give you the complete proof later for now, im just asking you a question..
Here it is again : suppose, you "know" that the given statement is true for \(n=k+1\), whenever it is true for \(n=k\). is that sufficient to claim that the given statement is true for all integers : \(n=1,2,3,\ldots\) ?
no?
why no ?
honestly idk
if you know that the statement is true for \(n=k+1\), whenever it is true for \(n=k\). that essentially means the statement is true for ALL positive integers if you could establish it for \(n=1\)
because, once you know that it is true for \(n=1\), it follows that it must be true for \(n=1+1=2\)
after you know that it is true for \(n=2\), it follows that it must be true for \(n=2+1=3\)
after you know that it is true for \(n=3\), it follows that it must be true for \(n=3+1=4\)
see the pattern ? it follows that it must be true for ALL integers greater than 1
ok i understand that
So all we need to do is to establish two things : 1) the statement is true for \(n=1\) 2) the statement is true for \(n=k+1\), whenever it is true for \(n=k\)
then by the previous induction patter, it will follow that the statement is true for ALL integers greater than 1
You have already proven that the statement is true for \(n=1\) so lets look at the second step
ok so what im doing is plugging in k+1?
Suppose the statement is true \(n=k\), then we have : \[8 + 16 + 24 + ... + 8k = 4k(k + 1)\]
add \(\color{red}{8(k+1)} \) both sides and get : \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4k(k + 1)\color{red}{+ 8(k+1)} \]
factoring out 4(k+1) on right hand side, do we get : \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4(k + 1)(k+2) \] ?
i'm sorry but all those parenthesis kinda confuses me.... O.O
thats okay, take ur time
i'm sorry but can you type it a bit more differently?.... is this what you're saying? 8+16+24+...+8k+8(k+1) = 4(k+1)(k+2)
Yes, is there any issue with latex ?
is it getting displayed like below on ur side ? https://i.gyazo.com/70406dea3fd312ded38a25e3f30da15b.png
its coming out like this: \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4(k + 1)(k+2) \]
take a screenshot and attach, browser messes up the latex sometimes
how do i screenshot it?
https://gyazo.com/
https://gyazo.com/e2eb27136301fb6cfe981d523de764f0
Ahh that looks horrible, I'll give you the screenshot of complete proof, one sec
Here is the complete proof https://i.gyazo.com/37b4e3e50a893db312121d7ed5de0944.png
let me knw if something doesn't make sense
oh ok i see.
i understand it. thank you alot.
yw

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