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

- anonymous

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- anonymous

@Shalante ?

- ganeshie8

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

- anonymous

yes

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## More answers

- ganeshie8

how, can u explain

- anonymous

4(1)(1+1)
4(2) =8

- ganeshie8

that means the expression \(4n(n+1)\) correctly gives the sum on left hand side for \(n=1\)

- ganeshie8

what if you have 2 terms on left hand side : \(8+16\)
is the statement still true ?

- anonymous

wait... you lost me. i have to plug in both now?

- ganeshie8

you don't need to, one base case is sufficient here, do you know how induction works ?

- anonymous

i'm a little rusty on induction.

- ganeshie8

basically your goal is to prove that the given statement is true for all positive integers : \(n=1,2,3,\ldots \)

- anonymous

i understand that. but I get confused after I've already proven 1

- anonymous

i don't know where to go from there

- ganeshie8

suppose, you "know" that the given statement is true for \(n=k+1\), whenever it is true for \(n=k\)

- ganeshie8

is that sufficient to claim that the given statement is true for all integers : \(n=1,2,3,\ldots\) ?

- anonymous

so then i have to plug in k+1

- ganeshie8

forget about the process, il surely give you the complete proof later
for now, im just asking you a question..

- 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

no?

- ganeshie8

why no ?

- anonymous

honestly idk

- 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

because, once you know that it is true for \(n=1\),
it follows that it must be true for \(n=1+1=2\)

- ganeshie8

after you know that it is true for \(n=2\),
it follows that it must be true for \(n=2+1=3\)

- ganeshie8

after you know that it is true for \(n=3\),
it follows that it must be true for \(n=3+1=4\)

- ganeshie8

see the pattern ?
it follows that it must be true for ALL integers greater than 1

- anonymous

ok i understand that

- 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

then by the previous induction patter, it will follow that the statement is true for ALL integers greater than 1

- ganeshie8

You have already proven that the statement is true for \(n=1\)
so lets look at the second step

- anonymous

ok so what im doing is plugging in k+1?

- ganeshie8

Suppose the statement is true \(n=k\), then we have :
\[8 + 16 + 24 + ... + 8k = 4k(k + 1)\]

- 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

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

i'm sorry but all those parenthesis kinda confuses me.... O.O

- ganeshie8

thats okay, take ur time

- 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

Yes, is there any issue with latex ?

- ganeshie8

is it getting displayed like below on ur side ?
https://i.gyazo.com/70406dea3fd312ded38a25e3f30da15b.png

- anonymous

its coming out like this: \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4(k + 1)(k+2) \]

- ganeshie8

take a screenshot and attach, browser messes up the latex sometimes

- anonymous

how do i screenshot it?

- ganeshie8

https://gyazo.com/

- anonymous

https://gyazo.com/e2eb27136301fb6cfe981d523de764f0

- ganeshie8

Ahh that looks horrible, I'll give you the screenshot of complete proof, one sec

- ganeshie8

Here is the complete proof
https://i.gyazo.com/37b4e3e50a893db312121d7ed5de0944.png

- ganeshie8

let me knw if something doesn't make sense

- anonymous

oh ok i see.

- anonymous

i understand it. thank you alot.

- ganeshie8

yw

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