The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

This discussion is now closed.

Check out other Related discussions

Proof by induction

prove the following is valid:

Unparseable latex formula:

a+ar+ar^2+...+ar^n^-^1 = a(\frac{1-r^n}{1-r})



ok so basically don't know at all where to start with this question never mind finish it...any pointers??
Reply 1
Avatar for rbnphlp
rbnphlp
11
adie_raz
prove the following is valid:

Unparseable latex formula:

a+ar+ar^2+...+ar^n^-^1 = a(\frac{1-r^n}{1-r})



ok so basically don't know at all where to start with this question never mind finish it...any pointers??

Let S=
Unparseable latex formula:

a+ar+ar^2+...+ar^n^-^1


multiply both sides by r
and find S .r

Spoiler


subtract S-Sr
Reply 2
Avatar for nuodai
nuodai
17
If you multiply the whole expression by rr then you get rS=ar+ar2+⋯+arnrS = ar + ar^2 + \cdots + ar^n. Notice that if you do S−rSS - rS all of the terms cancel except aa and arnar^n, leaving you with S−rS=a−arnS - rS = a - ar^n, so it's just a case of taking out common factors and dividing through.
Reply 3
Avatar for rbnphlp
rbnphlp
11
adie_raz
seriously have no clue how this helps at all :frown: i am sorry, really bad at this :confused:

Have you done what I have asked ?

Spoiler


Spoiler



when you get to S-Sr, take the S out S(1-r)
Reply 4
Avatar for nuodai
nuodai
17
I've just noticed that what me and rbnphlp said has nothing to do with induction :p:

What you need to do is to show that it's true for n=1 (i.e. a=a(1−r11−r)a = a \left( \dfrac{1-r^1}{1-r} \right), which is true). Assume that it's true for n=k, i.e. a+ar+⋯+ark−1=a(1−rk1−r)a + ar + \cdots + ar^{k-1} = a \left( \dfrac{1-r^k}{1-r} \right), and show that this implies that it's true for n=k+1 (try adding arkar^k to both sides of the equation).
Reply 5
Avatar for rbnphlp
rbnphlp
11
nuodai
I've just noticed that what me and rbnphlp said has nothing to do with induction :p:

What you need to do is to show that it's true for n=1 (i.e. a=a(1−r11−r)a = a \left( \dfrac{1-r^1}{1-r} \right), which is true). Assume that it's true for n=k, i.e. a+ar+⋯+ark−1=a(1−rk1−r)a + ar + \cdots + ar^{k-1} = a \left( \dfrac{1-r^k}{1-r} \right), and show that this implies that it's true for n=k+1 (try adding arkar^k to both sides of the equation).

:facepalm:
Reply 6
Avatar for adie_raz
adie_raz
OP
1
nuodai
I've just noticed that what me and rbnphlp said has nothing to do with induction :p:

What you need to do is to show that it's true for n=1 (i.e. a=a(1−r11−r)a = a \left( \dfrac{1-r^1}{1-r} \right), which is true). Assume that it's true for n=k, i.e. a+ar+⋯+ark−1=a(1−rk1−r)a + ar + \cdots + ar^{k-1} = a \left( \dfrac{1-r^k}{1-r} \right), and show that this implies that it's true for n=k+1 (try adding arkar^k to both sides of the equation).


this is what i was after thank you :P: i will give it a go and see how it goes, but dont count on the fact that i won't ask any more questions :o:
Reply 7
Avatar for adie_raz
adie_raz
OP
1
got it now thanks for the help guys :top:

Related discussions

Latest

Trending

Trending

Articles for you