Question

SOMEONE PLEASE HELP I'M ALMOST DONE.

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

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

Answer 1

@ParthKohli

Answer 2

True for the base case. Now assume that it is true for k, and then prove for k + 1.

Answer 3

Can you find the relationship between those two?

Answer 4

Is it supposed to be
\[\Large 4n(n+1)  \]
If that's the case, it must run a bit strangely.

\[n=1 \longrightarrow 8 \checkmark \\n=2 \longrightarrow24 \neg\checkmark \]

Answer 5

Actually, the order of operations first tell you to multiply 4 and \(n\), then \(n +1\)

Answer 6

I believe I can't follow you @ParthKohli

\[\Large 4n(n+1)=4n^2+4n \]
right? and still n=1 -> 8
n=2 -> 24,

or where do I miss something? (which is indeed possible=

Answer 7

The question is actually saying \((4 \cdot n) \cdot (n + 1)\)

Answer 8

But 8 + 16 = 24

Answer 9

It is a series, not a sequence.

Answer 10

oh I understand, this is where I misunderstood the problem, thank you.

Answer 11

No problem!

Answer 12

It's not explicit mentioned in the problem set, I have seen such problems as mainly sequences/progressions, in which this would be arithmetic.

Answer 13

Spacelimbus, can you please help the asker? I wish I could, but I am currently on a mobile device and it is hard for me to type.

Answer 14

Now that you know it works for the base case, you assume it holds for the case of "k", and you try to prove it works for "k + 1". So, add 8(k + 1) to both sides:

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

8 + 16 + 24 + ... + 8k + 8(k + 1) = 4k^2 + 4k + 8k + 8

8 + 16 + 24 + ... + 8k + 8(k + 1) = 4k^2 + 12k + 8

8 + 16 + 24 + ... + 8k + 8(k + 1) = 4(k^2 + 3k + 2)

8 + 16 + 24 + ... + 8k + 8(k + 1) = 4(k + 1)(k + 2)

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

But that is just the same expression we assumed true, but now holding for "k + 1", so this proves the case.

Answer 15

^ that is simple induction, correct.

Answer 16

All good now @Pssssst ?

Answer 17

The reason this proves the case is because of the way mathematical induction works. The base case has already been shown to be true. Our first "k" is "1" and we have proved it for "k + 1" or "2". "2" becomes our new "k", and by induction, it is proved for "k + 1" or "3", and this process goes on indefinitely. That is why mathematical induction works and why the above proof works.

Answer 18

Thank you, everyone! Helped a lot :)

Answer 19

Glad we were able to help. Thanks for the recognition!

Answer 20

@tcarroll010 Do you know much about history?