# Oxford PAT 2015, Question 3

## 6 thoughts on “Oxford PAT 2015, Question 3”

1. Teymour Beydoun says:

Hello,

Could you explain the 3rd line when you multiply the left part by 2? I do not understand how you get the part on the right of the equation.

Thank you

1. In the third line I’m not actually multiplying the LHS by 2. I’m subtracting Sn from both sides and then multiplying both sides by -1.

1. Teymour Beydoun says:

I see. Thanks a lot!

2. Hi.

Would you be able to explain the first 5 lines please? Or how you would go about evaluating a sum like this in general.

Thanks

1. Sure. In general, if you have to calculate a sum of values then try to find some operation on the sum that shifts its values right (or left) by one place. For example if summing a^i from i = 1 to i = 5, then it makes sense to consider multiplying the whole sum by a; this shifts the original sum a^1 + a^2 + a^3 + a^4 + a^5 one place to the right, giving us a^2 + a^3 + a^4 + a^5+ a^6, which is equal to the original sum minus the original leftmost term a^1, plus a newly-introduced rightmost term a^6. So if the original sum is S, then we can say that aS = S – a^1 + a^6.

That is exactly what we have done in the example: by considering what happens when we multiply the sum by 1/3 we find that (1/3)S = S – 1/3^1 + 1/3^6.

This kind of question seems to come up relatively often, and I think this approach to answering it is probably better than just trying to remember the formula for summing some geometric progression. If you make a mistake using the formula then it is easy for the answer to look as though you don’t have a clue what you are doing, but if you use the approach above there is less to remember (you just need to remember the general principle), and it should be obvious to the examiner what you are trying to do, so if you make a minor slip you are more likely to get two or three marks.