Solve (1+0.02/12)^12

$$( 1 + \frac { 0.02 } { 12 } ) ^ { 12 }$$

Evaluate

$$\frac{2220719284911033302994119543767201}{2176782336000000000000000000000000}\approx 1.020184356$$

Solution Steps

Expand \frac{0.02}{12} by multiplying both numerator and the denominator by 100.

$$\left(1+\frac{2}{1200}\right)^{12}$$

Reduce the fraction \frac{2}{1200} to lowest terms by extracting and canceling out 2.

$$\left(1+\frac{1}{600}\right)^{12}$$

Add 1 and \frac{1}{600} to get \frac{601}{600}.

$$\left(\frac{601}{600}\right)^{12}$$

Calculate \frac{601}{600} to the power of 12 and get \frac{2220719284911033302994119543767201}{2176782336000000000000000000000000}.

$$\frac{2220719284911033302994119543767201}{2176782336000000000000000000000000}$$

Factor

$$\frac{601 ^ {12}}{2 ^ {36} \cdot 3 ^ {12} \cdot 5 ^ {24}} = 1\frac{4.393694891103326 \times 10^{31}}{2.176782336 \times 10^{33}} = 1.0201843556815013$$