VCE Stuff > VCE Specialist Mathematics
Specialist 1/2 Question Thread!
Bri MT:
--- Quote from: Ruchir on December 07, 2020, 09:10:26 am ---Hello I am in year 11 and I am doing Spesh in the holidays , to get a head start.
I was stuck at this series question in the text book Ex 4E Q12
Find 1−x^2 +x^4 −x^6 +x^8 −···+x^2m ,where m is even.
Also please explain the answer.
Thanks :'(
--- End quote ---
Hey, I haven't done spec so I'm not 100% on how they would expect you to find it but here are some thoughts that may help:
This is an alternating series (it's a series where we see it alternating b/w + and -)
This then leads to thinking, well how do they get that alternation? We can do this by having (-1)^n and then when n is an odd number it will be (-1)^n = -1 but if n is an even number (-1)^2 = 1. (When you apply this double check if you start at positive or negative, sometimes you might do n + 1 rather than n as the power.)
Let's move on from the signs for a moment, we can easily see that we are incrementing the power of x by 2 each time. Since we start at a_0 = 1, we know that the power is starting at 0. We can represent this information using x^2n
Let's put this together:
a_n is the sum of (-1)^n x^2n
we are told that m is even (so that gives us a plus for our last term)
Hope this helps!
Ruchir:
--- Quote from: Bri MT on December 07, 2020, 12:00:28 pm ---Hey, I haven't done spec so I'm not 100% on how they would expect you to find it but here are some thoughts that may help:
This is an alternating series (it's a series where we see it alternating b/w + and -)
This then leads to thinking, well how do they get that alternation? We can do this by having (-1)^n and then when n is an odd number it will be (-1)^n = -1 but if n is an even number (-1)^2 = 1. (When you apply this double check if you start at positive or negative, sometimes you might do n + 1 rather than n as the power.)
Let's move on from the signs for a moment, we can easily see that we are incrementing the power of x by 2 each time. Since we start at a_0 = 1, we know that the power is starting at 0. We can represent this information using x^2n
Let's put this together:
a_n is the sum of (-1)^n x^2n
we are told that m is even (so that gives us a plus for our last term)
Hope this helps!
--- End quote ---
I appreciate the effort you took to solve this question but the answer is a bit different
This is the answer is a bit different.
keltingmeith:
--- Quote from: Ruchir on December 07, 2020, 09:10:26 am ---Hello I am in year 11 and I am doing Spesh in the holidays , to get a head start.
I was stuck at this series question in the text book Ex 4E Q12
Find 1−x^2 +x^4 −x^6 +x^8 −···+x^2m ,where m is even.
Also please explain the answer.
Thanks :'(
--- End quote ---
Have you covered the geometric series at all? If not, I reckon you just need to read up on that, and you'll probably be fine
Ruchir:
--- Quote from: keltingmeith on December 07, 2020, 01:35:03 pm ---
Have you covered the geometric series at all? If not, I reckon you just need to read up on that, and you'll probably be fine
--- End quote ---
Yes I have but this question is particular is baffling me, and I can’t see why there are supposed to be m+1 terms
keltingmeith:
--- Quote from: Ruchir on December 07, 2020, 01:49:49 pm ---Yes I have but this question is particular is baffling me, and I can’t see why there are supposed to be m+1 terms
--- End quote ---
Okay, well try breaking the series up and see if that makes things easier. Eg,
When m=1, you go up to x^(2m)=x^2: 1 - x^2
When m=4, you go up to x^(2m)=x^8: 1 - x^2 + x^4 - x^6 + x^8
Try a few more, fill out the series, count up the terms, and see what you notice
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