Solve (k+1)(k+2)(2k+2+1)/6

$$( k + 1 ) ( k + 2 ) ( 2 k + 2 + 1 ) / 6$$

Evaluate

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

Solution Steps

Add 2 and 1 to get 3.

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

Apply the distributive property by multiplying each term of k+1 by each term of k+2.

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

Combine 2k and k to get 3k.

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

Apply the distributive property by multiplying each term of k^{2}+3k+2 by each term of 2k+3.

$$\frac{2k^{3}+3k^{2}+6k^{2}+9k+4k+6}{6}$$

Combine 3k^{2} and 6k^{2} to get 9k^{2}.

$$\frac{2k^{3}+9k^{2}+9k+4k+6}{6}$$

Combine 9k and 4k to get 13k.

$$\frac{2k^{3}+9k^{2}+13k+6}{6}$$

Expand

$$\frac{k^{3}}{3}+\frac{3k^{2}}{2}+\frac{13k}{6}+1$$

Solution Steps

Add 2 and 1 to get 3.

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

Apply the distributive property by multiplying each term of k+1 by each term of k+2.

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

Combine 2k and k to get 3k.

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

Apply the distributive property by multiplying each term of k^{2}+3k+2 by each term of 2k+3.

$$\frac{2k^{3}+3k^{2}+6k^{2}+9k+4k+6}{6}$$

Combine 3k^{2} and 6k^{2} to get 9k^{2}.

$$\frac{2k^{3}+9k^{2}+9k+4k+6}{6}$$

Combine 9k and 4k to get 13k.

$$\frac{2k^{3}+9k^{2}+13k+6}{6}$$