ok so there's a GP

1 + (1+r) + (1+r)^2 + ....

a = 1 and r(common ratio) = (1+r) ??

Question

ok so there's a GP 

1 + (1+r) + (1+r)^2 + .... 

a = 1 and r(common ratio) = (1+r) ??

zugzwang · Answer

Seems like it... was there a further question?

blurbendy · Answer

(1 + r)^2 / (1 + r) = 1 + r
(1 + r ) / 1 = 1 + r
So, yes.

anonymous · Answer

well i had to find the sum, so i got r + 2, but i just wanted to check that I was going in the right direction.

zugzwang · Answer

This is only true if 
|1 + r| < 1
Let's assume it's positive, for the sake of simplicity...
0 < 1 + r < 1
-1 < r < 0

so the sum of the infinite geometric series is given by
$$\frac{a}{1-R}$$
Where a is the first term and R is the common ratio.  That gives
$$\frac{1}{1-(1+r)}$$

anonymous · Answer

ok, now in the next question, i don't know if its a continuation or not, so help :P

anonymous · Answer

On 1st Jan 2000, a man borrowed Rs. A at a fixed compound interest rate of a%. He pays back Rs. P on 1st Jan 2001, and then again Rs. P on 1st Jan 2002. Let a/100 = r 

Show that the amount still owing in Rs., just after this second payment, is A(1+r)^2 - P(1+r) - P.

zugzwang · Answer

I'm afraid this is beyond my scope :(

anonymous · Answer

awww.