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

$$k ( k + 1 ) \frac { ( 2 k + 1 ) + 6 ( k + 1 ) ^ { 2 } } { 6 }$$

Evaluate

$$\frac{k\left(k+1\right)\left(6k^{2}+14k+7\right)}{6}$$

Short Solution Steps

Express k\times \frac{2k+1+6\left(k+1\right)^{2}}{6} as a single fraction.

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

Use the distributive property to multiply \frac{k\left(2k+1+6\left(k+1\right)^{2}\right)}{6} by k+1.

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

Use the distributive property to multiply k by 2k+1+6\left(k+1\right)^{2}.

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

Express \frac{2k^{2}+k+6k\left(k+1\right)^{2}}{6}k as a single fraction.

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

Use the distributive property to multiply k by 2k+1+6\left(k+1\right)^{2}.

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

Since \frac{\left(2k^{2}+k+6k\left(k+1\right)^{2}\right)k}{6} and \frac{2k^{2}+k+6k\left(k+1\right)^{2}}{6} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(2k^{2}+k+6k\left(k+1\right)^{2}\right)k+2k^{2}+k+6k\left(k+1\right)^{2}.

$$\frac{2k^{3}+k^{2}+6k^{4}+12k^{3}+6k^{2}+2k^{2}+k+6k^{3}+12k^{2}+6k}{6}$$

Combine like terms in 2k^{3}+k^{2}+6k^{4}+12k^{3}+6k^{2}+2k^{2}+k+6k^{3}+12k^{2}+6k.

$$\frac{20k^{3}+21k^{2}+6k^{4}+7k}{6}$$

Expand

$$k^{4}+\frac{10k^{3}}{3}+\frac{7k^{2}}{2}+\frac{7k}{6}$$

Short Solution Steps

Express k\times \frac{2k+1+6\left(k+1\right)^{2}}{6} as a single fraction.

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

Use the distributive property to multiply \frac{k\left(2k+1+6\left(k+1\right)^{2}\right)}{6} by k+1.

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

Use the distributive property to multiply k by 2k+1+6\left(k+1\right)^{2}.

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

Express \frac{2k^{2}+k+6k\left(k+1\right)^{2}}{6}k as a single fraction.

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

Use the distributive property to multiply k by 2k+1+6\left(k+1\right)^{2}.

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

Since \frac{\left(2k^{2}+k+6k\left(k+1\right)^{2}\right)k}{6} and \frac{2k^{2}+k+6k\left(k+1\right)^{2}}{6} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(2k^{2}+k+6k\left(k+1\right)^{2}\right)k+2k^{2}+k+6k\left(k+1\right)^{2}.

$$\frac{2k^{3}+k^{2}+6k^{4}+12k^{3}+6k^{2}+2k^{2}+k+6k^{3}+12k^{2}+6k}{6}$$

Combine like terms in 2k^{3}+k^{2}+6k^{4}+12k^{3}+6k^{2}+2k^{2}+k+6k^{3}+12k^{2}+6k.

$$\frac{20k^{3}+21k^{2}+6k^{4}+7k}{6}$$