I'm really curious on how to find the answer to this question
John invests 100 at the end of year one, 200 at the end of year two, etc until he invests 1,000 at the end of year ten. The investment goes into a bank account earning 4%. At the end of each year, the interest is paid into a second bank account earning 3%. Calculate the total amount John will have after 10 years.
Here is what I did, I divided the two Funds to A and B.
Fund A has investments, as said in the problem
In fund B, if John reinvests the interest he earned on year 1 (from Fund A), then he will reinvest it on time 2 in Fund B in an amount of $$.04\cdot100 = 4$$
Likewise at time 3 in fund B, he will reinvest an amount of $$.04(100+200) = 12$$
etc.. which looks like
earning at an interest $j = 0.03$
As far as I know, the increase in Fund B is not of arithmetic progression or geometric. Hence, when I accumulate Fund B, I'm somewhat using a "brute force" method $$AV_B(t=10) = 4(1.03)^8+12(1.03)^7+\cdot\cdot\cdot 144(1.03)+180$$ $$= 701.32$$
Hence, since Fund A has total of $100+200+..+1000 = 5500$, then I know that the accumulated value for both fund is $$5500 + 701.32 = 6201.32$$
as opposed to the book's answer $6200.52$ which is, in my pursuit of "exact answer", mine is somewhat close. (I'm studying for exam FM, and I'm really attentive to the "exact" answer).
My question is, is there any more "beautiful" approach to this problem as opposed to what I did? Or is my approach correct at all? Thank you for your answers
$$S = .04\times 100\left[\dfrac{1.03^9-1}{.03}+2.\dfrac{1.03^8-1}{.03}+3.\dfrac{1.03^7-1}{.03}+\cdots+8.\dfrac{1.03^2-1}{.03}+9\right]$$
$$ S = .04\times 100\left(\dfrac{1.03^9}{0.03} + 2.\dfrac{1.03^8}{0.03} + \cdots 8.\dfrac{1.03^2}{0.03}\right) -.04\times 100\left(\frac{1}{0.03} (1+2+..+8)-9\right)$$
$$ S_1 = .04\times 100\frac{1.03^2}{0.03} \left( 8+7.1.03+6.1.03^2+\cdots+1.03^7\right) $$
https://en.wikipedia.org/wiki/Arithmetico%E2%80%93geometric_sequence
$$ S_2 = \dfrac{a-(a+(n-1)d)r^n}{1-r} + \dfrac{dr(1-r^{n-1})}{(1-r)^2}$$
a = 8, d = -1, r = 1.03
$$S_2 = \dfrac{8-(8+(8-1)(-1))1.03^8}{1-1.03} + \dfrac{(-1)1.03(1-1.03^7)}{(1-1.03)^2}$$
$$=-224.441 + 263.078 = 38.637$$
$$ S_1= .04\times 100\frac{1.03^2}{0.03}\left(38.637\right)=1366.33389$$
Thus
$$S = 4\times(1366.33389-1191) = 701.34$$