$${2(Ia)20-a20-20^2v^(21)\over v(1-v)}$$

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Hi there :)

Your answer is slightly off.

This is the way I did it:

$$PV = 1 + 4v + 9v^2 + ... + 400v^{19} $$

$$vPV = v + 4v^2 + 9v^3 + ... + 400v^{20} $$

Now we need to take \(PV - vPV\) and we get:

$$PV(1-v) = (1 + 3v + 5v^2 + ... + 39v^{19}) - 400v^{20}$$

Then we stick with first principles and make the terms in brackets into a level annuity and an increasing annuity.

$$PV(1-v) = (1 + v + v^2 + ... + v^{19}) + 2(v + 2v^2 + 3v^3 + ... + 19v^{19} ) - 400 v^{20} $$

$$= \ddot{a}_\bar{n|} + 2(Ia)_\bar{n|} - 400v^{20} $$

Therefore:

$$PV = \frac{\ddot{a}_\bar{n|} + 2(Ia)_\bar{n|} - 400v^{20} }{1-v}$$

commented
Aug 16, 2016
by
michwairish
(400 points)

Thank you, I understand

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