egbeach
  • egbeach
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. 1. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) =
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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geerky42
  • geerky42
4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) \(\textbf=\) ??? what?
egbeach
  • egbeach
it doesnt say... thats how it was written
zzr0ck3r
  • zzr0ck3r
ouch

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zzr0ck3r
  • zzr0ck3r
it is not a statement...
geerky42
  • geerky42
we cannot prove expression "4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2)"
geerky42
  • geerky42
I mean it doesn't make sense.
egbeach
  • egbeach
thats what i thought
zzr0ck3r
  • zzr0ck3r
I cant find the formula elsewhere.
geerky42
  • geerky42
What I realize though is that pattern is supposed to be \(n(n+2)\) starting at \(n=4\), not \(4n(4n+2)\) Because otherwise pattern is supposed to be \(4\cdot6+8\cdot10+12\cdot14+\cdots\)
egbeach
  • egbeach
okay i found it. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = (4(4n+1)(8n+7))/6
zzr0ck3r
  • zzr0ck3r
right @geerky42
zzr0ck3r
  • zzr0ck3r
Do you want to see if it matches the formula she found?
geerky42
  • geerky42
Maybe I misinterpreted the pattern, it's just \(\displaystyle \sum_{i=4}^{4n}i(i+2)\) ?
egbeach
  • egbeach
wait i found the answer since the statement is false. thank you though

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