Question

Use mathematical induction to prove the statement is true for all positive integers n.

8 + 16 + 24 + ... + 8n = 4n(n + 1)

Answer 1

@Shalante ?

Answer 2

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

Answer 3

yes

Answer 4

how, can u explain

Answer 5

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

Answer 6

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

Answer 7

what if you have 2 terms on left hand side : \(8+16\)

is the statement still true ?

Answer 8

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

Answer 9

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

Answer 10

i'm a little rusty on induction.

Answer 11

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

Answer 12

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

Answer 13

i don't know where to go from there

Answer 14

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

Answer 15

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

Answer 16

so then i have to plug in k+1

Answer 17

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

Answer 18

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 : \(n=1,2,3,\ldots\) ?

Answer 19

no?

Answer 20

why no ?

Answer 21

honestly idk

Answer 22

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 \(n=1\)

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

ok i understand that

Answer 28

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 \(n=k\)

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

add \(\color{red}{8(k+1)} \) both sides and get :

\[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4k(k + 1)\color{red}{+ 8(k+1)} \]

Answer 34

factoring out 4(k+1) on right hand side, do we get :

\[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4(k + 1)(k+2) \]

?

Answer 35

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

Answer 36

thats okay, take ur time

Answer 37

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)

Answer 38

Yes, is there any issue with latex ?

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

how do i screenshot it?

Answer 43

https://gyazo.com/

Answer 44

https://gyazo.com/e2eb27136301fb6cfe981d523de764f0

Answer 45

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

Answer 46

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

Answer 47

let me knw if something doesn't make sense

Answer 48

oh ok i see.

Answer 49

i understand it. thank you alot.

Answer 50

yw