HSC Stuff > HSC Mathematics Extension 1
3U Maths Question Thread
RuiAce:
--- Quote from: milie10 on February 12, 2020, 12:33:27 am ---Thanks Rui, I did it!! Is this right? I'm in the process of learning a bit of latex so its a bit misaligned haha :)
derivative:
--- End quote ---
Yeah it looks correct. Wolfram agrees with your answer as well. :)
shekhar.patel:
Hi. I just need some suggestion with this question and my working out for it.
I have got the answer attached below, but the marker's answer is 4(k+2).
So is my answer correct and if so, are there many ways to get the same answer, because the marker used the difference between the (k+1)th and k th equation to get 4(k+2). Thanks
RuiAce:
--- Quote from: shekhar.patel on February 16, 2020, 12:44:57 pm ---Hi. I just need some suggestion with this question and my working out for it.
I have got the answer attached below, but the marker's answer is 4(k+2).
So is my answer correct and if so, are there many ways to get the same answer, because the marker used the difference between the (k+1)th and k th equation to get 4(k+2). Thanks
--- End quote ---
You haven't specified the question in full. Are you trying to prove that \(n(n+2)\) is divisible by 4 only for the even integers?
shekhar.patel:
--- Quote from: RuiAce on February 16, 2020, 06:34:38 pm ---You haven't specified the question in full. Are you trying to prove that \(n(n+2)\) is divisible by 4 only for the even integers?
--- End quote ---
Yes for even integers
fun_jirachi:
Hey there!
Induction for this question is actually quite unnecessary - note for some even integer n, it can be expressed as 2k (for some integer k), and as such we have that 2k(2k+2) = 4(k)(k+1) which is an integer divisible by 4, since k and k+1 are clearly integers. Therefore, I would doubt that this question would be presented in an exam - out of curiosity, where's this question from?
However, if you insist on using induction, I will show you the inductive step:
Assuming that for n=k, \(k(k+2) = 4a \ \ (k, a \in \mathbb{Z}^+)\)
Then for n=k+2:
Hope this helps :)
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