Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. 4 x 6 + 5 x 7 + 6 x 8 +...+ 4n(4n+2) = 4(4n+1)(8n+7)/6
First put it in the sum of a series formula.
\[\sum_{r=1}^{n}4n(4n+2)\]
Although that doesn't appear to be correct because that would give you 4x6 + 8x10 + 12x14......
@phi Any input?
what is the value for n=1 ?
I mean, what is the sum of a series formula here?
Oh, sorry just saw this. But if I plug in n =1, then I get 2
I am completely lost, so will you please walk me through it from step one?
I would have said it was\[\sum_{r=1}^{n}(r+3)(r+5)\]
for n=1 there is only one term in the summation 4n(4n+2) --> 4*6 which is 24 however 4(4n+1)(8n+7)/6 with n=1 is not 24 so we are not doing too well...
Well it says to show why it is false... So I just will say that it is false and therefore cannot be proved true.
4n(4n+2) is incorrect isn't it?? Surely it's (n+3)(n+5)?
4n(4n+2) is not 5x7 when n=2.
Or 6x8 when n=3.
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. In an induction proof, you start by proving a "base case" in this problem, the base case (for n=1) is not true, so the formula is proven false at the first step.
Okay so now I plug in 1 everywhere were n is.. getting 50
So, I show my work for n=1, then state that it is false?
yes, and the series has only one term which clearly is 24 24 ≠ 50 therefore, false
Okay, thank you!
Are you just ignoring everything i'm saying?
Noo! I just didn't understand what you meant.
No, I meant phi :)
This series says that the last term is (4n)(4n+2). I'm saying that that can't be right because all of the previous terms were (n+3)(n+5). If n is 1 you get 4x6. If n is 2 you get 5x7. If n is 3 you get 6x7. If you use 4n(4n+2), that doesn't work with the other terms.
I'm going to post a new question and close this one.. unless you two would like to help.
& OHHH,ok.
Because if n=2 then you get 8x10. If n=3 then you get 12x14.
@phi Is what I've said not correct?
@alrightatmaths probably, but it doesn't make the formula correct.
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