With regards to this same question, is this the answer to part (a) of the question

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$${2(Ia)20-a20-20^2v^(21)\over v(1-v)}$$
related to an answer for: Increasing Annuity

answered Aug 16, 2016 by (1,490 points)
selected Aug 18, 2016

Hi there :)

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 (400 points)
Thank you, I understand