A deposit of $2000 is made in an account that earns 6% interest compounded monthly. The balance in the account after n months is given by $A _ { n } = 2000 \left( 1 + \frac { 0.06 } { 12 } \right) ^ { n } , n = 1,2,3 , \mathrm {~K}$ Find the balance in the account after 11 years by finding the 132th term of the sequence. Round to the nearest penny.
A)$4,379,295.09
B)$3863.23
C)$3844.01
D)$265,320.00
E)$3882.54

Question

Quizplus · Accepted Answer

The balance after 11 years (or 132 months) is calculated using the formula $A _ { n } = 2000 \left( 1 + \frac { 0.06 } { 12 } \right) ^ { 132 }$. Plugging in the values, the correct balance is $3863.23, rounded to the nearest penny.

A Deposit of $2000 Is Made in an Account That An=2000(1+0.0612)n,n=1,2,3, KA _ { n } = 2000 \left( 1 + \frac { 0.06 } { 12 } \right) ^ { n } , n = 1,2,3 , \mathrm {~K}An​=2000(1+120.06​)n,n=1,2,3, K

A Deposit of $2000 Is Made in an Account That $A _ { n } = 2000 \left( 1 + \frac { 0.06 } { 12 } \right) ^ { n } , n = 1,2,3 , \mathrm {~K}$