Deriving formulas for the present value of a series of regular payments

[@klaus7602]

I really admire people who is good at math, ty.

[@victordeprez]

5:04 you told me to pee on both sides of the equation, but it didn’t solve my problem

[@anirbandas2054]

Thank you,you saved my test

[@karamnaddour1768]

Thank you for simplifying this.

[@toroscanislupus]

thank you very much. It is very helpfull. Hello from Turkey:)

[@jiayanglu6449]

excellent video!

[@infiniteera-c8t]

5:00 -sign ??

[@hamzaelfadli6715]

@@infiniteera-c8t it was used to cancel out the extra number(the last number) that P has.

[@ishwaranandkabra]

Great job dude

[@xxaqpunyaxx]

thank you so muchh !! you've been a great help !

[@jiayanglu6449]

I have a question though. I tried to come up with the formula for future value of annuity using your method 1. However, I just can't get it right. I've arrived at P[(1+r)/1 -1]= A [1 - (1+r)^(n-1)] and then try to solve for P. I tried multiple times to simplify the current version but none of the derivations gets me the most simplified version. Could you explain it, please? Do I have to use geometric sequence to derive the formula for FV of annuity? Thank you so much for your help Professor.

[@chuckloner]

Great video!

[@onakoyafaruk262]

Can I get a video for future value?

[@DarshanGhorpade7]

Let P(t) denote the present value (at the time 0) of the amount 1 that is to be received at the time t. Show that r(t) is a nondecreasing function of t if and only if P(αt) ≥ (P(t))^α

[@infiniteera-c8t]

5:00 where did this - sign came from ....?

[@mathslearningUofA]

The top line has P = A/(1+i) + ... + A/(1+i)^n. If I subtract A/(1+i)^n from both parts of this equation I get P - A/(1+i)^n = A/(1+i) + ...+ A/(1+i)^{n-1}. Thus when I see A/(1+i) + ...+ A/(1+i)^{n-1} in the second line, I can replace it with P - A/(1+i)^n.

[@dianafarhat9479]

Thanks!

[@lebronpeps6]

Thanks man!

[@hermanwolf4024]

Yup. I'm still an idiot. 🙄🔫