Verified answer

sum 100(1.05^x*1.07^-(x+1)), x = 0 to 9 + sum 100*1.05^9*0.95^y*1.07^-(y+10), y = 1 to 9

= $1270.06

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The primary quandary you've gotten is what do you imply through 11 1/2% curiosity is that each year powerful? Is that a nominal cost? In which case you can must understand what the compounding interval is as is possibly eleven half of% each year compounded monthly. You'll be able to want to figure out what the monthly interest cost is and if that 11 1/2% is a real powerful every year expense then you could possibly take the twelfth root of 1.One hundred fifteen to offer you 1.009112468 which is 0.9112468% per 30 days. If it is eleven half% per annum compounded month-to-month then it could be 0.958333% per 30 days. The 2nd factor that you're lacking is if the $325 is being deposited at the starting or the top of the month. After that it's a easy summation of a geometrical sequence quandary which is truely taught in junior excessive however nobody appears to bear in mind of that. If we count on that the interest cost is eleven half% per annum nominal compounded month-to-month as a consequence a zero.958333% per month curiosity rate and we let R = 1 + zero.958333 / one hundred or 1.00958333 which is also 1 + 11.5/1200. We let X be the $325 monthly deposit and we assume that the deposit is on the commencing of the month then we can say the following: After one month, the balance is: F(1) = X * R After two months, the steadiness is: F(2) = ( F(1) + X ) * R .: F(2) = X * R^2 + X * R After three months, the balance is: F(three) = ( F(2) + X ) * R F(three) = X * R^three + X * R^2 + X * R As you'll find this may also be generalized as: F(n) = X * summation of the time period R^ok for okay = 1 to n applying the summation of a geometrical sequence equation, this turns into: F(n) = X * ( ( 1 - R^(n+1) ) / ( 1 - R ) - 1 ) there's 312 months from his thirty-ninth to sixty-fifth birthday so we've a future price of: F(312) = $325 * ( ( 1 - ( 1 + 11.5/1200 )^313 ) / ( 1 - ( 1 + eleven.5/1200 ) ) - 1 ) F(312) = $636,984.Eighty one word that if the deposits have been at the end of the month, it really works out to be: F(n) = X * summation of the time period R^ok for k = 0 to n - 1 .: F(n) = X * ( 1 - R^n ) / ( 1 - R ) F(312) = $325 * ( 1 - ( 1 + eleven.5/1200)^312 ) / ( 1 - ( 1 + eleven.5/1200 ) ) F(312) = $630,938.32 And of direction, the numbers would even be specific if that 11.5% was an strong cost or a nominal fee with one more compounding interval equivalent to compounded continuously. So that you quite ought to get the query correct in terms of what the interest cost relatively is and when the deposits are without a doubt being made.

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