Michael is saving money and plans on making monthly contributions into an account earning an annual interest rate of 5.7\% compounded monthly. If Michael would like to end up with $\$ 137,000$ after 10 years, how much does he need to contribute to the account every month, to the nearest dollar? Use the following formula to determine your answer.
\[
A=d\left(\frac{(1+i)^{n}-1}{i}\right)
\]
$A=$ the future value of the account after $n$ periods
$d=$ the amount invested at the end of each period
$i=$ the interest rate per period
$n=$ the number of periods

Step 1 ：Given that Michael wants to end up with $137,000 after 10 years, and the account earns an annual interest rate of 5.7% compounded monthly, we need to find out how much he needs to contribute to the account every month.

Step 2 ：We use the formula for the future value of a series of monthly contributions, which is given by \[A=d\left(\frac{(1+i)^{n}-1}{i}\right)\] where \(A\) is the future value of the account after \(n\) periods, \(d\) is the amount invested at the end of each period, \(i\) is the interest rate per period, and \(n\) is the number of periods.

Step 3 ：We know the values of \(A\), \(i\), and \(n\). \(A\) is $137,000, \(i\) is the monthly interest rate which is the annual interest rate divided by 12 (5.7% / 12), and \(n\) is the number of periods which is 10 years times 12 (10 * 12).

Step 4 ：We can rearrange the formula to solve for \(d\): \[d=\frac{A*i}{(1+i)^{n}-1}\]

Step 5 ：Substituting the known values into the formula, we get \(d = \frac{137000*0.00475}{(1+0.00475)^{120}-1}\)

Step 6 ：Calculating the above expression, we find that \(d\) is approximately 849.6735017025178

Step 7 ：Rounding to the nearest dollar, we find that Michael needs to contribute approximately $850 to the account every month.

Step 8 ：So, the final answer is \(\boxed{850}\)