The Student Room Group

FP1 Summation Question Help (OCR)

Hello everyone. I've come across a particular question in the Summation chapter of the OCR Further Pure 1 textbook, and I was wondering if anyone could write out a solution to PART 2 for me as I have not been able to get anything equivalent to the answer in the back of the textbook (Part 1 is ok, Part 2 is the problem). The question is Question 7, from Exercise 2C, Page 35, I've copied it down in the images below.

Part 1:
CodeCogsEqn.gif
Part 2:
CodeCogsEqn (1).gif
The answer given in the back of the textbook:
CodeCogsEqn (2).gif

Thanks a lot (to whoever replies)! :smile:
Note that

a) from part 1

(rar(r1)ar1)=(a1)rar1+ar1 \sum (ra^r - (r-1)a^{r-1}) = (a-1)\sum ra^{r-1} + \sum a^{r-1}

b) the LHS telescopes down to something trivial
c) the first term on the RHS is (almost) what you want to find
d) the second term on the RHS is the sum of a GP
Reply 2
Thankyou very much! I was using a long-winded method which is why it was not fully working, but this makes more sense! :smile:


Posted from TSR Mobile
Original post by holatombola
Thankyou very much! I was using a long-winded method which is why it was not fully working, but this makes more sense! :smile:


Posted from TSR Mobile


You can also approach the problem directly:

S=1nrar1=1+2a+3a2+4a3++(n1)an2+nan1 \displaystyle S = \sum_1^n ra^{r-1} = 1+ 2a + 3a^2 + 4a^3 + \cdots + (n-1)a^{n-2} + na^{n-1}

So:

aS=a+2a2+3a3+4a4++(n1)an1+nanaS = a + 2a^2 + 3a^3 + 4a^4 + \cdots + (n-1)a^{n-1} + na^n

Hence:

SaS=(1a)S=1+a+a2++an1nan=an1a1nanS-aS = (1-a)S = 1 + a +a^2 + \cdots + a^{n-1} - na^n = \frac{a^n-1}{a-1}-na^n

and so on.

Quick Reply

Latest