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

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Is the statement true for \(n=1\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0that means the expression \(4n(n+1)\) correctly gives the sum on left hand side for \(n=1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what if you have 2 terms on left hand side : \(8+16\) is the statement still true ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait... you lost me. i have to plug in both now?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0you don't need to, one base case is sufficient here, do you know how induction works ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm a little rusty on induction.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0basically your goal is to prove that the given statement is true for all positive integers : \(n=1,2,3,\ldots \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i understand that. but I get confused after I've already proven 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't know where to go from there

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0suppose, you "know" that the given statement is true for \(n=k+1\), whenever it is true for \(n=k\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0is that sufficient to claim that the given statement is true for all integers : \(n=1,2,3,\ldots\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so then i have to plug in k+1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0forget about the process, il surely give you the complete proof later for now, im just asking you a question..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Here 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 : \(n=1,2,3,\ldots\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0if 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
 one year ago
Best ResponseYou've already chosen the best response.0because, once you know that it is true for \(n=1\), it follows that it must be true for \(n=1+1=2\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0after you know that it is true for \(n=2\), it follows that it must be true for \(n=2+1=3\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0after you know that it is true for \(n=3\), it follows that it must be true for \(n=3+1=4\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0see the pattern ? it follows that it must be true for ALL integers greater than 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i understand that

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0then by the previous induction patter, it will follow that the statement is true for ALL integers greater than 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0You have already proven that the statement is true for \(n=1\) so lets look at the second step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so what im doing is plugging in k+1?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Suppose the statement is true \(n=k\), then we have : \[8 + 16 + 24 + ... + 8k = 4k(k + 1)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0add \(\color{red}{8(k+1)} \) both sides and get : \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4k(k + 1)\color{red}{+ 8(k+1)} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0factoring 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
 one year ago
Best ResponseYou've already chosen the best response.0i'm sorry but all those parenthesis kinda confuses me.... O.O

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0thats okay, take ur time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'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
 one year ago
Best ResponseYou've already chosen the best response.0Yes, is there any issue with latex ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0is it getting displayed like below on ur side ? https://i.gyazo.com/70406dea3fd312ded38a25e3f30da15b.png

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its coming out like this: \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4(k + 1)(k+2) \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0take a screenshot and attach, browser messes up the latex sometimes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do i screenshot it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Ahh that looks horrible, I'll give you the screenshot of complete proof, one sec

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Here is the complete proof https://i.gyazo.com/37b4e3e50a893db312121d7ed5de0944.png

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0let me knw if something doesn't make sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i understand it. thank you alot.
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