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) =

Question

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) =

geerky42 · Answer

4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) $\textbf=$ ??? what?

egbeach · Answer

it doesnt say... thats how it was written

zzr0ck3r · Answer

ouch

zzr0ck3r · Answer

it is not a statement...

geerky42 · Answer

we cannot prove expression "4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2)"

geerky42 · Answer

I mean it doesn't make sense.

egbeach · Answer

thats what i thought

zzr0ck3r · Answer

I cant find the formula elsewhere.

geerky42 · Answer

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 · Answer

okay i found it. 
4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = (4(4n+1)(8n+7))/6

zzr0ck3r · Answer

right @geerky42

zzr0ck3r · Answer

Do you want to see if it matches the formula she found?

geerky42 · Answer

Maybe I misinterpreted the pattern, it's just $\displaystyle \sum_{i=4}^{4n}i(i+2)$ ?

egbeach · Answer

wait i found the answer since the statement is false. thank you though