Hello,

Could someone please explain why you would do the answer to (ii) subtracted from (iii)a to get the answer for (iii)b?

I do not understand how this works. I've also attached what I thought the answer would be. For reference, this is question 9 from Jan 2006 FP1 OCR paper.

Could someone please explain why you would do the answer to (ii) subtracted from (iii)a to get the answer for (iii)b?

I do not understand how this works. I've also attached what I thought the answer would be. For reference, this is question 9 from Jan 2006 FP1 OCR paper.

(edited 10 months ago)

Just writing it out as a simple series you have

sum from n+1 to inf = sum from 1 to inf - sum from 1 to n

or equivalently as per the previous comment

sum from 1 to n + sum from n+1 to inf = sum from 1 to inf

Its one way to derive the sum to n formula of a geometric series, based on the infinite sum. The numerator is a - ar^(n+1) which represents

sum from 1 to inf - sum from n+1 to inf

as ar^(n+1) is the first term in the geometric series that starts at n+1.

sum from n+1 to inf = sum from 1 to inf - sum from 1 to n

or equivalently as per the previous comment

sum from 1 to n + sum from n+1 to inf = sum from 1 to inf

Its one way to derive the sum to n formula of a geometric series, based on the infinite sum. The numerator is a - ar^(n+1) which represents

sum from 1 to inf - sum from n+1 to inf

as ar^(n+1) is the first term in the geometric series that starts at n+1.

(edited 10 months ago)

Original post by mqb2766

Just writing it out as a simple series you have

sum from n+1 to inf = sum from 1 to inf - sum from 1 to n

or equivalently as per the previous comment

sum from 1 to n + sum from n+1 to inf = sum from 1 to inf

Its one way to derive the sum to n formula of a geometric series, based on the infinite sum. The numerator is a - ar^(n+1) which represents

sum from 1 to inf - sum from n+1 to inf

as ar^(n+1) is the first term in the geometric series that starts at n+1.

sum from n+1 to inf = sum from 1 to inf - sum from 1 to n

or equivalently as per the previous comment

sum from 1 to n + sum from n+1 to inf = sum from 1 to inf

Its one way to derive the sum to n formula of a geometric series, based on the infinite sum. The numerator is a - ar^(n+1) which represents

sum from 1 to inf - sum from n+1 to inf

as ar^(n+1) is the first term in the geometric series that starts at n+1.

So sum from n+1 to inf = sum from 1 to inf - sum from 1 to n is always true for any summations?

Original post by action123

So sum from n+1 to inf = sum from 1 to inf - sum from 1 to n is always true for any summations?

You could have written down a simple example but f the series was the sum of natural numbers (ignoring the fact that the infinite sum doesnt converge...) then the sum from 1 to infinity is

1 + 2 + 3 +....+ n + (n+1) + (n+2) + ...

Obviously you can write this as

[1 + 2 + 3+....+n] + [(n+1) + (n+2) + ... ]

so

sum from 1 to n + sum from (n+1) to inf = sum from 1 to inf

You shouldnt have to remember it, it drops out of the definition of the series (splitting at n/n+1). Assuming the sum to infinity converges, then its true for any series.

(edited 10 months ago)

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