anonymous
  • anonymous
Use mathematical induction to prove the statement is true for all positive integers n. 8 + 16 + 24 + ... + 8n = 4n(n + 1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@Shalante ?
ganeshie8
  • ganeshie8
Is the statement true for \(n=1\) ?
anonymous
  • anonymous
yes

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ganeshie8
  • ganeshie8
how, can u explain
anonymous
  • anonymous
4(1)(1+1) 4(2) =8
ganeshie8
  • ganeshie8
that means the expression \(4n(n+1)\) correctly gives the sum on left hand side for \(n=1\)
ganeshie8
  • ganeshie8
what if you have 2 terms on left hand side : \(8+16\) is the statement still true ?
anonymous
  • anonymous
wait... you lost me. i have to plug in both now?
ganeshie8
  • ganeshie8
you don't need to, one base case is sufficient here, do you know how induction works ?
anonymous
  • anonymous
i'm a little rusty on induction.
ganeshie8
  • ganeshie8
basically your goal is to prove that the given statement is true for all positive integers : \(n=1,2,3,\ldots \)
anonymous
  • anonymous
i understand that. but I get confused after I've already proven 1
anonymous
  • anonymous
i don't know where to go from there
ganeshie8
  • ganeshie8
suppose, you "know" that the given statement is true for \(n=k+1\), whenever it is true for \(n=k\)
ganeshie8
  • ganeshie8
is that sufficient to claim that the given statement is true for all integers : \(n=1,2,3,\ldots\) ?
anonymous
  • anonymous
so then i have to plug in k+1
ganeshie8
  • ganeshie8
forget about the process, il surely give you the complete proof later for now, im just asking you a question..
ganeshie8
  • ganeshie8
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 : \(n=1,2,3,\ldots\) ?
anonymous
  • anonymous
no?
ganeshie8
  • ganeshie8
why no ?
anonymous
  • anonymous
honestly idk
ganeshie8
  • ganeshie8
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 \(n=1\)
ganeshie8
  • ganeshie8
because, once you know that it is true for \(n=1\), it follows that it must be true for \(n=1+1=2\)
ganeshie8
  • ganeshie8
after you know that it is true for \(n=2\), it follows that it must be true for \(n=2+1=3\)
ganeshie8
  • ganeshie8
after you know that it is true for \(n=3\), it follows that it must be true for \(n=3+1=4\)
ganeshie8
  • ganeshie8
see the pattern ? it follows that it must be true for ALL integers greater than 1
anonymous
  • anonymous
ok i understand that
ganeshie8
  • ganeshie8
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 \(n=k\)
ganeshie8
  • ganeshie8
then by the previous induction patter, it will follow that the statement is true for ALL integers greater than 1
ganeshie8
  • ganeshie8
You have already proven that the statement is true for \(n=1\) so lets look at the second step
anonymous
  • anonymous
ok so what im doing is plugging in k+1?
ganeshie8
  • ganeshie8
Suppose the statement is true \(n=k\), then we have : \[8 + 16 + 24 + ... + 8k = 4k(k + 1)\]
ganeshie8
  • ganeshie8
add \(\color{red}{8(k+1)} \) both sides and get : \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4k(k + 1)\color{red}{+ 8(k+1)} \]
ganeshie8
  • ganeshie8
factoring out 4(k+1) on right hand side, do we get : \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4(k + 1)(k+2) \] ?
anonymous
  • anonymous
i'm sorry but all those parenthesis kinda confuses me.... O.O
ganeshie8
  • ganeshie8
thats okay, take ur time
anonymous
  • anonymous
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)
ganeshie8
  • ganeshie8
Yes, is there any issue with latex ?
ganeshie8
  • ganeshie8
is it getting displayed like below on ur side ? https://i.gyazo.com/70406dea3fd312ded38a25e3f30da15b.png
anonymous
  • anonymous
its coming out like this: \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4(k + 1)(k+2) \]
ganeshie8
  • ganeshie8
take a screenshot and attach, browser messes up the latex sometimes
anonymous
  • anonymous
how do i screenshot it?
ganeshie8
  • ganeshie8
https://gyazo.com/
anonymous
  • anonymous
https://gyazo.com/e2eb27136301fb6cfe981d523de764f0
ganeshie8
  • ganeshie8
Ahh that looks horrible, I'll give you the screenshot of complete proof, one sec
ganeshie8
  • ganeshie8
Here is the complete proof https://i.gyazo.com/37b4e3e50a893db312121d7ed5de0944.png
ganeshie8
  • ganeshie8
let me knw if something doesn't make sense
anonymous
  • anonymous
oh ok i see.
anonymous
  • anonymous
i understand it. thank you alot.
ganeshie8
  • ganeshie8
yw

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