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)

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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)

Mathematics
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True for the base case. Now assume that it is true for k, and then prove for k + 1.
Can you find the relationship between those two?

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Is it supposed to be \[\Large 4n(n+1)  \] If that's the case, it must run a bit strangely. \[n=1 \longrightarrow 8 \checkmark \\n=2 \longrightarrow24 \neg\checkmark \]
Actually, the order of operations first tell you to multiply 4 and \(n\), then \(n +1\)
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=
The question is actually saying \((4 \cdot n) \cdot (n + 1)\)
But 8 + 16 = 24
It is a series, not a sequence.
oh I understand, this is where I misunderstood the problem, thank you.
No problem!
It's not explicit mentioned in the problem set, I have seen such problems as mainly sequences/progressions, in which this would be arithmetic.
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.
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.
^ that is simple induction, correct.
All good now @Pssssst ?
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.
Thank you, everyone! Helped a lot :)
Glad we were able to help. Thanks for the recognition!
@tcarroll010 Do you know much about history?

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