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As soon as she graduated from college, Kay began planning for her retirement. Her plans were to deposit $500 semiannually into an IRA (a retirement fund) beginning 6 months after r graduation and continuing until the day she retired, which she expected to be 30 years later. Today is the day Kay retires. She just made the last $500 deposit into her retirement fund, and now she wants to know how much she has accumulated for her retirement. The fund earned 10 percent compounded semiannually since it was established.

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a. Compute the balance of the retirement fund assuming all the payments were made on time. b. Although Kay was able to make all of the $500 deposits she planned, 10 years ago she had to withdraw $ I0000 from the fund to pay some medical bills incurred by her mother. Compute the balance in the retirement fund based on this information.
lets see, I got no idea if the "6 months after graduation" affects anything. 10%/2 = 5% The semiannully means twice a year. and 30 years at 2 times a year = 60 cycles altogether. Beginning balance = B{0} = 500; B{n} = B{n-1} (1.05) + 500 B{n} tracks the balance per cycle; B{n-1} refers to the previous balance 1.05 is the interest added to the Balance 500 is the extra deposit each time B{0} = 500 B{1} = 1025 B{2} = 1576.25 B{3} = 2155.0625 B{4} = 2762.815.... While this method works, it can be quite laborious over large cycles. We can devise a formula that is based on the number of the cycle itself that will tell us the balance for any given cycle. We simply start with the general equation and manipulate it to get back to our starting point of B{1} B{n} = B{n-1} (1.05) + 500 The Balance previous, the B{n-1} was figured in the same fashion as the one we are currently working on. B{n-1} = B{n-2}(1.05) + 500 ; since we know what B{n -1} is we can substitute it into the general equation to get: B{n} = [B{n-2}(1.05) + 500](1.05) + 500 = B{n-2}(1.05)^2 + 500(1.05) + 500 This recursive quality can continue forever, but by doing this backstepping; we can see a pattern forming that can be used to make an actual formula to use. B{n-2} = B{n-3}(1.05) + 500 ; insert it and simplify B{n} = [B{n-3}](1.05) + 500](1.05)^2 + 500(1.05) + 500 = B{n-3}(1.05)^3 + 500(1.05)^2 + 500(1.05) + 500 We have 2 distinct area forming; B{n-r}(1.05)^r and: 500(1.05)^r-1 + 500(1.05)^r-2 + ... + 500(1.05) + 500 Our goal is to get to B{0}; B{n-r} = B{0} when: n - r = 0, or simply put; when r = n. B{n-n} = B{0} B{0}(1.05)^n 500(1.05)^n-1 + 500(1.05)^n-2 + ... + 500(1.05) + 500 The first bit is easy; its just stays as is and we need to make that infinitely growing last part to come into focus now. Lets factor out the 500 to clear our view: 500[ (1.05)^n-1 +(1.05)^n-2+ ... +(1.05)^2 +(1.05) +1 ] Lets also rename it to see its behaviour better and maybe tame it some; 500(A) Now our issue is with "A", let work on it... A = {(1.05)^n-1 +(1.05)^n-2+ ... +(1.05)^2 +(1.05) +1} If we multiply A by 1.05; we can shift all of its innards; for example: (1.05)(1.05)^n-1 = (1.05)^n (1.05)(1.05)^n-2 = (1.05)^n-1 .... etc The trick is to subtract these 2 sets of values and see if that helps :) \(\small\color{green}{(1.05)A = [(1.05)^n +(1.05)^{n-1}+(1.05)^{n-2}+ ... +(1.05)^2 +(1.05) \ \ \ \ \ \ \ ]}\) \(\small\color{red}{ -(1.00)A = [}\) \(\small\color{red}{(1.05)^{n-1} +(1.05)^{n-2}+ ... +(1.05)^2 +(1.05) +1]}\) ---------------------------------------------------- \(\small\color{blue}{(0.05)A=[(1.05)^n.................................................-1]}\) Now we solve for "A" :) (.05)A = 1.05^n -1 ; divide out the .05 A = \(\color{purple}{\frac{1.05^n-1}{.05}}\) -------------------------------------- Now that weve cleaned it up; lets use it :) \(B{n} = B{0}(1.05)^n +500(A)\) \( = 500(1.05)^n +500(\frac{1.05^n-1}{.05})\) \(=500\left(1.05^n+\frac{1.05^n-1}{.05} \right)\) \(=500\left(\frac{(.05)1.05^n+1.05^n-1}{.05} \right)\) \(=\frac{500}{.05}[(.05)1.05^n+1.05^n-1]\) \(=10000[(.05)1.05^n+1.05^n-1]\) \(=(500)1.05^n+(10000)1.05^n-10000\) \(=1.05^n(500+10000)-10000\) \(=1.05^n(10500)-10000\) Thats the simplest form I can think of, lol And when we plug in n=60 we get: B{60} = 10,500(1.05^(60))-10,000 = 186,131.45.....
part b: suggests that after 10 years prior to the end (30-10=20) they withdrew 10,000 which would suggest that we need to compute (20*2=40) cycles and subtract 10,000 from it to calibrate our stuff. B{40} - 10,000 becomes our new beginning balance point and we can work it out from there in the same manner :) Im gonna introduce a new variable and call it "p"; this will indicate the number of cycles AFTER they withdrew the 10,000. \[B_{40}+p=(B_{40}āˆ’10,000)(1.05)^p+(500)\frac{1.05^pāˆ’1}{.05}\] This when p=20; we are at 60 cycles (30 years) and it should give a good balance amount

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Other answers:

it comes to abt 159,598.4748.... if i did it right :)
if the 30 yrs is from the date of graduation, then i believe n should be 59, since it begins 6 months from graduation. google annuities for formula

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