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Fred invests an amount of money in an account paying r% compound interest per annum. the amount of money doubles after 'n' years. find a formula for 'r' in terms of 'n' r=
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Let A be the initial amount invested and B be the current about. After every year, B increases by Br/100 so, after the first year, B=A+Ar/100=A(1+r/100) After the second year, B increases by another Br/100 -> B=A+Ar/100+(A+Ar/100)r/100=A+2Ar/100+A(r/100)^2=A(1+r/100)^2 After year m, B=A(1+r/100)^m At year n, B=2A so : \[2=(1+r/100)^n\implies2^{-n}=1+r/100\implies100(2^{-n}-1)=r\]
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