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

yes

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.

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

no?

why no ?

honestly idk

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

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 :
\[8 + 16 + 24 + ... + 8k = 4k(k + 1)\]

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

thats okay, take ur time

Yes, is there any issue with latex ?

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