Solve 1+-2+-2^2/2+-2^3/6

$$1+-2+ \frac{ { -2 }^{ 2 } }{ 2 } + \frac{ { -2 }^{ 3 } }{ 6 }$$

Evaluate

$$-\frac{1}{3}\approx -0.333333333$$

Solution Steps

Subtract 2 from 1 to get -1.

$$-1+\frac{\left(-2\right)^{2}}{2}+\frac{\left(-2\right)^{3}}{6}$$

Calculate -2 to the power of 2 and get 4.

$$-1+\frac{4}{2}+\frac{\left(-2\right)^{3}}{6}$$

Divide 4 by 2 to get 2.

$$-1+2+\frac{\left(-2\right)^{3}}{6}$$

Add -1 and 2 to get 1.

$$1+\frac{\left(-2\right)^{3}}{6}$$

Calculate -2 to the power of 3 and get -8.

$$1+\frac{-8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

$$1-\frac{4}{3}$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{3}{3}-\frac{4}{3}$$

Since \frac{3}{3} and \frac{4}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3-4}{3}$$

Subtract 4 from 3 to get -1.

$$-\frac{1}{3}$$

Factor

$$-\frac{1}{3} = -0.3333333333333333$$