anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@ParthKohli
ParthKohli
  • ParthKohli
True for the base case. Now assume that it is true for k, and then prove for k + 1.
ParthKohli
  • ParthKohli
Can you find the relationship between those two?

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

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