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egbeach
 one year ago
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) =
egbeach
 one year ago
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) =

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geerky42
 one year ago
Best ResponseYou've already chosen the best response.04 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) \(\textbf=\) ??? what?

egbeach
 one year ago
Best ResponseYou've already chosen the best response.0it doesnt say... thats how it was written

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0it is not a statement...

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0we cannot prove expression "4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2)"

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0I mean it doesn't make sense.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I cant find the formula elsewhere.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0What 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
 one year ago
Best ResponseYou've already chosen the best response.0okay i found it. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = (4(4n+1)(8n+7))/6

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Do you want to see if it matches the formula she found?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0Maybe I misinterpreted the pattern, it's just \(\displaystyle \sum_{i=4}^{4n}i(i+2)\) ?

egbeach
 one year ago
Best ResponseYou've already chosen the best response.0wait i found the answer since the statement is false. thank you though
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