Solve (k+1)^2(k+2)^2/4(2k+1)+k/24[2k^2+9k+13]

$$\frac { ( k + 1 ) ^ { 2 } ( k + 2 ) ^ { 2 } } { 4 ( 2 k + 1 ) } + \frac { k } { 24 } [ 2 k ^ { 2 } + 9 k + 13 ]$$

Evaluate

$$\frac{10k^{4}+56k^{3}+113k^{2}+85k+24}{24\left(2k+1\right)}$$

Short Solution Steps

Use the distributive property to multiply 4 by 2k+1.

$$\frac{\left(k+1\right)^{2}\left(k+2\right)^{2}}{8k+4}+\frac{k}{24}\left(2k^{2}+9k+13\right)$$

Express \frac{k}{24}\left(2k^{2}+9k+13\right) as a single fraction.

$$\frac{\left(k+1\right)^{2}\left(k+2\right)^{2}}{8k+4}+\frac{k\left(2k^{2}+9k+13\right)}{24}$$

Factor 8k+4.

$$\frac{\left(k+1\right)^{2}\left(k+2\right)^{2}}{4\left(2k+1\right)}+\frac{k\left(2k^{2}+9k+13\right)}{24}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(2k+1\right) and 24 is 24\left(2k+1\right). Multiply \frac{\left(k+1\right)^{2}\left(k+2\right)^{2}}{4\left(2k+1\right)} times \frac{6}{6}. Multiply \frac{k\left(2k^{2}+9k+13\right)}{24} times \frac{2k+1}{2k+1}.

$$\frac{6\left(k+1\right)^{2}\left(k+2\right)^{2}}{24\left(2k+1\right)}+\frac{k\left(2k^{2}+9k+13\right)\left(2k+1\right)}{24\left(2k+1\right)}$$

Since \frac{6\left(k+1\right)^{2}\left(k+2\right)^{2}}{24\left(2k+1\right)} and \frac{k\left(2k^{2}+9k+13\right)\left(2k+1\right)}{24\left(2k+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{6\left(k+1\right)^{2}\left(k+2\right)^{2}+k\left(2k^{2}+9k+13\right)\left(2k+1\right)}{24\left(2k+1\right)}$$

Do the multiplications in 6\left(k+1\right)^{2}\left(k+2\right)^{2}+k\left(2k^{2}+9k+13\right)\left(2k+1\right).

$$\frac{6k^{4}+24k^{3}+24k^{2}+12k^{3}+48k^{2}+48k+6k^{2}+24k+24+4k^{4}+2k^{3}+18k^{3}+9k^{2}+26k^{2}+13k}{24\left(2k+1\right)}$$

Combine like terms in 6k^{4}+24k^{3}+24k^{2}+12k^{3}+48k^{2}+48k+6k^{2}+24k+24+4k^{4}+2k^{3}+18k^{3}+9k^{2}+26k^{2}+13k.

$$\frac{10k^{4}+56k^{3}+113k^{2}+85k+24}{24\left(2k+1\right)}$$

Expand 24\left(2k+1\right).

$$\frac{10k^{4}+56k^{3}+113k^{2}+85k+24}{48k+24}$$

Expand

$$\frac{10k^{4}+56k^{3}+113k^{2}+85k+24}{24\left(2k+1\right)}$$

Short Solution Steps

Use the distributive property to multiply 4 by 2k+1.

$$\frac{\left(k+1\right)^{2}\left(k+2\right)^{2}}{8k+4}+\frac{k}{24}\left(2k^{2}+9k+13\right)$$

Express \frac{k}{24}\left(2k^{2}+9k+13\right) as a single fraction.

$$\frac{\left(k+1\right)^{2}\left(k+2\right)^{2}}{8k+4}+\frac{k\left(2k^{2}+9k+13\right)}{24}$$

Factor 8k+4.

$$\frac{\left(k+1\right)^{2}\left(k+2\right)^{2}}{4\left(2k+1\right)}+\frac{k\left(2k^{2}+9k+13\right)}{24}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(2k+1\right) and 24 is 24\left(2k+1\right). Multiply \frac{\left(k+1\right)^{2}\left(k+2\right)^{2}}{4\left(2k+1\right)} times \frac{6}{6}. Multiply \frac{k\left(2k^{2}+9k+13\right)}{24} times \frac{2k+1}{2k+1}.

$$\frac{6\left(k+1\right)^{2}\left(k+2\right)^{2}}{24\left(2k+1\right)}+\frac{k\left(2k^{2}+9k+13\right)\left(2k+1\right)}{24\left(2k+1\right)}$$

Since \frac{6\left(k+1\right)^{2}\left(k+2\right)^{2}}{24\left(2k+1\right)} and \frac{k\left(2k^{2}+9k+13\right)\left(2k+1\right)}{24\left(2k+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{6\left(k+1\right)^{2}\left(k+2\right)^{2}+k\left(2k^{2}+9k+13\right)\left(2k+1\right)}{24\left(2k+1\right)}$$

Do the multiplications in 6\left(k+1\right)^{2}\left(k+2\right)^{2}+k\left(2k^{2}+9k+13\right)\left(2k+1\right).

$$\frac{6k^{4}+24k^{3}+24k^{2}+12k^{3}+48k^{2}+48k+6k^{2}+24k+24+4k^{4}+2k^{3}+18k^{3}+9k^{2}+26k^{2}+13k}{24\left(2k+1\right)}$$

Combine like terms in 6k^{4}+24k^{3}+24k^{2}+12k^{3}+48k^{2}+48k+6k^{2}+24k+24+4k^{4}+2k^{3}+18k^{3}+9k^{2}+26k^{2}+13k.

$$\frac{10k^{4}+56k^{3}+113k^{2}+85k+24}{24\left(2k+1\right)}$$

Expand 24\left(2k+1\right).

$$\frac{10k^{4}+56k^{3}+113k^{2}+85k+24}{48k+24}$$