Question

let a denote the first term and r the common ratio of an exponential sequence.
(i) Show that the general rule for finding the sum, Sn' of the fist n terms of the sequence is given by S_n=(a(1-r^n))/(1-r),r≠1.
(ii) Hence deduce the sum to infinity of an exponential squence with |r|<1.

Answer 1

Write S_n and then find r*S_n
See if you can see a pattern in them! You will be surprised

Answer 2

Hmm, I don't understand

Answer 3

S_n = a + a*r + a*r^2 + a*r^3 + a*r^4.............nterms
Do you understand this?

Answer 4

yes, is that the starting

Answer 5

yeah.......now write r*S_n

Answer 6

ok

Answer 7

rSn=ar+〖ar〗^2+〖ar〗^3+⋯+〖ar〗^(n-1)+〖ar〗^n

Answer 8

Ah great...........can you see most of the terms in S_n and r*S_n are same

Answer 9

yes

Answer 10

great..now guess what will happen when you subtract them

Answer 11

you mean Sn-rSn

Answer 12

yeah!!

Answer 13

a-ar^n

Answer 14

So S -rS = a - ar^n right??
I am sure you can find S now

Answer 15

good

Answer 16

what of (ii)

Answer 17

oh it is very similar!! there will be infinite terms for both S and r*S.........so you need not care about the last terms

Answer 18

Try and write it out

Answer 19

ok trying now

Answer 20

hmm, no idea

Answer 21

what does infinity mean??

Answer 22

no ending

Answer 23

yeah exactly .....Infinity means that 'continues for ever'

Answer 24

So if I write S for infinity terms
S = a + ar + ar^2 + ar^3 ..............continues for ever

Answer 25

and then I can write rS = ar + ar^2 + ar^3 + .............continues for ever

Answer 26

Now what will happen if I subtract S and rS

Answer 27

you are left with a

Answer 28

exactly
so S-rS =a
u can now find S for sure

Answer 29

S is 1

Answer 30

oh no!!

Answer 31

S-rS =a
so, S(1-r) =a (I have taken S common)
Now find S

Answer 32

S=a/(1-r)

Answer 33

is this correct?

Answer 34

yeah right!!!

Answer 35

Wow, atlas

Answer 36

Thanks but I have questions

Answer 37

more questions

Answer 38

yeah sure go on