#series

35 messages · Page 1 of 1 (latest)

solemn hawkBOT
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sharp panther
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First of all, you need the sums of individual terms:
Σ(1, r = 1 to n) = n
Σ(r, r = 1 to n) = n(n + 1)/2
Σ(r^2, r = 1 to n) = n(n + 1)(2n + 1)/6
Σ(a^r, r = 1 to n) = a(a^n - 1)/(a - 1)
In the last term you can bring k outside of the sum, as it's a constant.

red path
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I don’t know the 4th one

sharp panther
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We have:
Σ(3r^2 + 8r + 3 + k 2^(r - 1), r = 1 to 12) = 3250
First, let's use the linearity of sum:
3Σ(r^2, r = 1 to 12) + 8Σ(r, r = 1 to 12) + 3Σ(1, r = 1 to 12) + kΣ(2^(r - 1), r = 1 to 12) = 3250
So, now you can use the formulas above to evaluate all the sums, then solve for k.

sharp panther
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You didn't calculate the sums, though. Just their last term for some reason.

red path
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Not in the Edexcel fm textbook

sharp panther
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Ah, actually, no, you did, but I don't know how exactly you did it.

red path
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I didn’t calculate the others because they were in the previous part of the q

sharp panther
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Oh, alright.

red path
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Did

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Here’s the mark scheme which shows the formula you said

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But what’s the -art that says adds up all 12 terms how would I do that

sharp panther
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Well, no need to manually add them up. There is a formula for the sum of geometric progression.

red path
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But how would you add up manually?

sharp panther
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Well, just add each term to the previous one.

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Obviously, this isn't very rational.

red path
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But that still requires that formula?

sharp panther
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No, the formula allows you to calculate it by just using it, without needing to take a long time to add the terms one by one.

red path
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But how would I do that long method

sharp panther
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Well, you would just do 1 + 2 + 4 + ... + 2^11.

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But why would you do that?

red path
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We haven’t been taught gp yet

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Why would it start at 2^0 not 1?

sharp panther
red path
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Oh shit I’m stupid

sharp panther
red path
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Thanks buddy have a good night

red path
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Especially further maths

sharp panther
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Hm. Well, alright.

sharp panther