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Jason deposits $3960$ into a bank account at $t=0$. The bank credits interest at the end of each year at a force of interest

$\delta_t=\frac{1}{8+t}$

Interest can be reinvested at an annual effective rate of $7$%. The total accumulated amount at $t=3$ is equal to X. Calculate X.

I started finding $i$

$e^{\delta}=1+i$

Replacing $t=1$, $i=0.1175$

Hence $X=3960$X$0.001175(\frac{1.07^3-1}{0.07})+3960=5456.14$

But the answer is $5393.53$

Tosh
  • 1,614

1 Answers1

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Let be $K=3960$. The accumulation function is $$a(t)=\mathrm e^{\int_{0}^t\delta(s)\mathrm d s}=\mathrm e^{\int_{0}^t\frac{1}{8+s}\mathrm d s}=\mathrm e^{\log(8+s)\big|_0^t}=\mathrm e^{\log\left(\frac{8+t}{8}\right)}=\frac{8+t}{8}=1+\frac{t}{8}$$ The interest earned in the first year is $$I_1=K\frac{a(1)-a(0)}{a(0)}=3960\cdot\frac{1}{8}=495$$ The interest earned in the second year is $$I_2=K\frac{a(2)-a(1)}{a(1)}=3960\cdot\frac{1}{9}=440$$ The interest earned in the third year is $$I_3=K\frac{a(3)-a(2)}{a(2)}=3960\cdot\frac{1}{10}=396$$ Since the 3 interest payements are accumulated at rate of $i=7\%$, the accumulated amount of money $A$ at time $t=3$ is \begin{align} A&=K+I_1(1+i)^2+I_2(1+i)+I_3\\ &=3960+495(1.07)^2+440(1.07)+396 \\ &=3960 + 566.72 + 470.8 + 396 \\ &= 5393.53 \end{align}

alexjo
  • 14,976