The amount of money in Nadia's savings account at the end of n years is represented by the recursive formula a_(n)=1.025(a_(n-1))+50 If a_(6)=291.51 , what is the balance in Nadia's account at the end of each of the first two years? 25.36, 75.99 50.00, 101.25 101.25, 153.78 75.99, 127.89

To find the balance in Nadia's account at the end of each of the first two years, we can use the recursive formula given:<br /><br />$a_{n}=1.025(a_{n-1})+50$<br /><br />We are given that $a_{6}=291.51$. We can use this information to find the balance at the end of each year from year 1 to year 6.<br /><br />Let's start with year 1:<br /><br />$a_{1}=1.025(a_{0})+50$<br /><br />Since we don't know the initial amount in Nadia's account, we'll assume it to be $a_{0}$.<br /><br />Now let's move on to year 2:<br /><br />$a_{2}=1.025(a_{1})+50$<br /><br />We can substitute the value of $a_{1}$ from the equation above:<br /><br />$a_{2}=1.025(1.025(a_{0})+50)+50$<br /><br />Simplifying this equation, we get:<br /><br />$a_{2}=1.025^{2}(a_{0})+1.025(50)+50$<br /><br />Now we need to find the value of $a_{0}$ in order to calculate $a_{2}$. We can use the given information that $a_{6}=291.51$:<br /><br />$291.51=1.025^{6}(a_{0})+1.025^{5}(50)+...+1.025(50)+50$<br /><br />Simplifying this equation, we get:<br /><br />$291.51=a_{0}(1.025^{6})+50(1.025^{5})+...+50(1.025)+50$<br /><br />Now we can solve for $a_{0}$ using a calculator or computer program. Once we have the value of $a_{0}$, we can substitute it back into the equations for $a_{1}$ and $a_{2}$ to find the balance in Nadia's account at the end of each of the first two years.<br /><br />After solving the equation, we find that the balance in Nadia's account at the end of each of the first two years is $\$ 50.00,\$ 101.25$.