Edexcel Further Maths May 2025 CP1 Q5: Method of Differences (Summation Proof)
Comprehensive worked solution and examiner mark-scheme analysis by Tutor Sheefa. Focuses on mark allocation, algebraic traps, and step-by-step mathematical reasoning.
The Exam Archetype: Telescoping Series via Partial Fractions
Examiners do not invent new mathematics each year — they test consistent structural archetypes. In this paper, Q5 tests your ability to navigate the boundary between conceptual algebra and precise numerical computation.
$$\sum_{r=1}^n [f(r) - f(r+1)] = f(1) - f(n+1)$$
$$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$$
Examiner Traps & Common Mark-Scheme Penalties
- Writing '...' without displaying enough initial and final terms to clearly demonstrate the cancellation pattern.
- Algebraic errors when combining surviving fractions into a single simplified expression.
Formal Mathematical Solution
Below is the full step-by-step derivation meeting official mark-scheme criteria for method (M) and accuracy (A) marks:
Displaying Telescoping Pattern
Clearly show the two initial uncancelled terms and two final uncancelled terms.
🎙️ Read Spoken Video Explanation (239 segments, 1286 words) ▾ Expand
Unedited transcript of Tutor Sheefa's spoken audio instructions during the walkthrough:
Number five, we have a series question and they want us to use method of difference to prove that it can be rewritten in that form. Okay, let's begin. We have four over r squared minus one. Let's first express this in partial fraction. In order to express this in partial fraction, we have to rewrite our denominator in factored form. This is actually r squared minus one can be rewrite rewritten as r squared minus one squared. Difference of two square. That means the factored form is r minus one r plus one. So four over r minus one r plus one is equals to a over r minus one plus b over r plus one. So [snorts] this is our partial fraction. How can we find or how can we figure out the value of a and b? We have to first multiply this equation by the factors. So r minus one r plus one. So what happened when we multiply by this? We will get this. So here would be r minus one r plus one r minus one r plus one. Okay, let me erase this part. So why do we choose this? Because we want to eliminate the fraction. >> [snorts] >> There we go. We have four equals to a r plus one plus b r minus one. Next, we need to find two values of r that we can substitute into equation one so that we can find the value of a and b. Let's first begin by finding each root. Okay. r plus one equal to zero, that means r equals to negative one. r minus one equal to zero, that means r equals to one. These two values should be substituted into this equation. So let r equals to negative one into one. So four equals to a negative one plus one, that is zero plus b. Negative one minus one, that's negative two. So negative two b is equals to four. B is equals to negative two. We're done. Let's move on to the second value. And let r equals to one into equation one. Four is equal to a one plus one is two. Plus b one minus one is zero. Therefore, we have two a is equal to four. A is equal to two. Let's rewrite our partial fraction. So we have our series r from two until n four over r squared minus one. This is equal to r squared equal r equals to two until end. Okay, let's begin with a. Okay, a two over r minus one. And [snorts] then we have negative two, that means we can put minus here. And then two up here. Okay, so this is more the series sign. So from two until end, we have negative. Okay, here would be r plus one. Let's separate. Let's do uh the series separately. Let's expand them. Let's begin with the orange one. Let's first substitute two. So two over two minus one plus two over three minus one plus two over four minus one. Okay, we have three. And then we can just put dot dot dot and the last three. What are the last three? It would be n minus two, n minus one, and n. So two over n minus two minus one plus two over n minus one minus one plus two over n minus one. We are done with the first expansion. Let's continue with the second one. This is from the olive series. So two over two plus one plus two over three plus one plus two over four plus one. Just three would be enough and then you put dot dot dot and let's do the last three, which is the same thing. So two over n minus two plus one plus two over n minus one plus one plus two over n [snorts] plus one. That's it. Let's expand and simplify the numbers. So I will get, okay, two over one plus two over two plus two over three plus dot dot dot plus two over n minus three plus two over n minus two plus two over n minus one. Minus two over three plus two over four plus two over five plus dot dot dot plus two over n minus one plus two over n plus two over n plus one. Okay. Now let's see, we have some values that we can add on in the middle to make sure that when we cancel the terms, uh it will be easier to cancel the terms. Okay. Let me continue with our computation up here. [snorts] Okay. So now we know that the numbers here right? Actually add up by what? It keep increasing by one. So what I can do is I can add on. I know for sure after two over three, I have two over four and two over five. And [snorts] that's it. I can continue with the plus dot dot dot. Okay. And then the same thing here. If I reverse, I know that before n minus one is n minus two. And before n minus two, I would have n minus three. That's the reason why I put it like, you know, um the orange series at the top and the olive series below. That is to make it easier for you to cancel the terms. Okay, [snorts] let's start canceling. Two over three can be canceled by two over three. Two over four, two over four. Two over five, two over five. That's it. Notice that I can cancel this with this. Now we can just write down all the balance terms I have. Two plus one minus two over n minus two over n plus one. Why is these two minus? Because I distribute the negative sign here. See this negative. So when I distribute, both of the terms would be negative as well. Okay. So here I have three over one minus two over n minus 2 over n + 1. I want to combine these three fractions so that it becomes one. So, I have to multiply the first fraction with n n + 1. And then the second one just n + 1. And then for the third one, I have to multiply by n. Okay. That's it. Let's expand the numerator. We will have 3 n squared + 3 n 2 n - 2 2 n over n n + 1. So, here it will be 3 n squared minus 3 n - 2 n minus Okay, that will be minus n - 2 n n + 1. Next, we have to factor our numerator. We can simply use our calculator. Okay, go to equation polynomial degree two. Okay, coefficient is three -1 -2. Okay, the first the first value of n is 1. And the second one is -2 over 3. We want to find factor. So, when we want to find factor, we have to make it equal to zero. So, n - 1 equal to zero. Here, you have to bring the three up first. So, 3 n equals to -2. 3 n + 2 equal to zero. Now, both of these are equal to zero. So, we know this is the first factor and this is the second one. So, that's it. The first one is n - 1. And the second one is 3 n + 2. Okay. Equals to n n + 1. We can now erase this. We do not need to show that. Okay. With that, we are done. We managed to prove that it looks like that. So, this is our answer for question number five. That's it.