Solve (k+1)(k+2)/2-k(k+1)/2

$$\frac { ( k + 1 ) ( k + 2 ) } { 2 } - \frac { k ( k + 1 ) } { 2 }$$

Evaluate

$$k+1$$

Short Solution Steps

Since \frac{\left(k+1\right)\left(k+2\right)}{2} and \frac{k\left(k+1\right)}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(k+1\right)\left(k+2\right)-k\left(k+1\right)}{2}$$

Do the multiplications in \left(k+1\right)\left(k+2\right)-k\left(k+1\right).

$$\frac{k^{2}+2k+k+2-k^{2}-k}{2}$$

Combine like terms in k^{2}+2k+k+2-k^{2}-k.

$$\frac{2k+2}{2}$$

Divide each term of 2k+2 by 2 to get 1+k.

$$1+k$$

Expand

$$k+1$$

Short Solution Steps

Since \frac{\left(k+1\right)\left(k+2\right)}{2} and \frac{k\left(k+1\right)}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(k+1\right)\left(k+2\right)-k\left(k+1\right)}{2}$$

Do the multiplications in \left(k+1\right)\left(k+2\right)-k\left(k+1\right).

$$\frac{k^{2}+2k+k+2-k^{2}-k}{2}$$

Combine like terms in k^{2}+2k+k+2-k^{2}-k.

$$\frac{2k+2}{2}$$

Divide each term of 2k+2 by 2 to get 1+k.

$$1+k$$