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anonymous
 3 years ago
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)
anonymous
 3 years ago
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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ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1True for the base case. Now assume that it is true for k, and then prove for k + 1.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Can you find the relationship between those two?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is it supposed to be \[\Large 4n(n+1) \] If that's the case, it must run a bit strangely. \[n=1 \longrightarrow 8 \checkmark \\n=2 \longrightarrow24 \neg\checkmark \]

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Actually, the order of operations first tell you to multiply 4 and \(n\), then \(n +1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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=

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1The question is actually saying \((4 \cdot n) \cdot (n + 1)\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1It is a series, not a sequence.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh I understand, this is where I misunderstood the problem, thank you.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's not explicit mentioned in the problem set, I have seen such problems as mainly sequences/progressions, in which this would be arithmetic.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Spacelimbus, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now 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.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1^ that is simple induction, correct.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0All good now @Pssssst ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you, everyone! Helped a lot :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Glad we were able to help. Thanks for the recognition!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@tcarroll010 Do you know much about history?
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