deriving the formula for decreasing annuity

[asmkdas]

Can anyone please help me to derive (Da)_n=(n-a_n)/i

 

[Tim.Sullivan]

Good question. I have had a quick browse and can't find it, so I have attempted it myself, below. I'd be grateful for feedback from others!

Equation 1 DA_N = nV +(n-1)V^2 +(n-2)V^3......+3V^n-2 +2V^n-1 +V^n

Multiply these through by (1+i) to give:

Equation2 (1+i) DA_N = n +(n-1)V +(n-2)V^2......+3V^n-3 +2V^n-2 +V^n-1

Subtract : Equation 2 - Equation 1 gives

i x DA_N = n -V -V^2........-V^n-2 -V^n-1 -V^n
or
i x DA_N = n -(v + v^2.....+v^n-2 +v^n-1 +v^n)

Since the formula in the brackets is the formula for a level annuity in arrears:

i x DA_N = n-a_n

And so DA_N = (n-a_n)/i

 

[asmkdas]

a new sum

Good question. I have had a quick browse and can't find it, so I have attempted it myself, below. I'd be grateful for feedback from others!

Equation 1 DA_N = nV -(n-1)V^2 -(n-2)V^3......-3V^n-2 -2V^n-1 -V^n

Multiply these through by (1+i) to give:

Equation2 (1+i) DA_N = n -(n-1)V -(n-2)V^2......-3V^n-3 -2V^n-2 -V^n-1

Subtract : Equation 2 - Equation 1 gives

i x DA_N = n -V -V^2........-V^n-2 -V^n-1 -V^n
or
i x DA_N = n -(v + v^2.....+v^n-2 +v^n-1 +v^n)

Since the formula in the brackets is the formula for a level annuity in arrears:

i x DA_N = n-a_n

And so DA_N = (n-a_n)/i

Thank you very much, though I had done it in different way like,
(Da)_n=[(n+1)a_n]-(Ia)_n
=[(n+1)(1-v^n)/i]-[(1-v^n)/(i/(1+i))-nv^n]/i
=i^-1[in-1+v^n]
=[n-a_n]/i

Again I would like to thank you for sharing me another method.

I got a question, like:
Find the Present Value of a series of payments viz. 50 at time 1, 45 at time 2, 40 at time 3...upto 6 times.

We can find the answer as:
PV=[(55a_6)-5(Ia)_6]@3%=205.478588

But I cant get the same answer when I am using (Da)_n=(n-a_n)/i

Please help.

 

[Tim.Sullivan]

Find the Present Value of a series of payments viz. 50 at time 1, 45 at time 2, 40 at time 3...upto 6 times. We can find the answer as:
PV=[(55a_6)-5(Ia)_6]@3%=205.478588

In the example given they use:
a) 55a_6 = 55V + 55V^2 + 55v^3 + 55v^4 + 55v^5 + 55v^6
b) -5Ia_6= -5V - 10V^2 - 15v^3 - 20v^4 - 25v^5 - 30v^6
Combined = 50v + 45V^2 + 40v^3 +35 v^4 + 30v^5 + 25v^6

Using the DA_N calc we need:
a) 20a_6 = 20V + 20V^2 + 20v^3 + 20v^4 + 20v^5 + 20v^6
b) 5Da_6= 30V + 25V^2 + 20v^3 + 15v^4 + 10v^5 + 5v^6
combined = 50v + 45V^2 + 40v^3 +35 v^4 + 30v^5 + 25v^6

So (55a_6 -5Ia_6) = (20a_6 +5Da_6)
I've checked this using the formula Da_6 = (6-a_6)/0.03

This has helped me realise that my orginal formula below was slightly wrong (I've edited it now to make it correct) and to prove it here:

Equation 1: Da_6 = 6v + 5v^2 + 4v^3 + 3v^4 + 2v^5 + v^6
Multiply through by (1+i)
Equation 2: (1+i)Da_6 = 6 + 5v + 4v^2 + 3v^3 + 2v^4 + v^5
Subtracting: Equation 2 - Equation 1
Gives: iDa_6 = 6 -v -v^2 -v^3 -v^4 -v^5 -v^6
so iDa_6 = 6 - a_6
Da_6 = (6 - a_6)/i