Solve 1/3+1/2n+1-1/2n+3

$$\frac { 1 } { 3 } + \frac { 1 } { 2 n + 1 } - \frac { 1 } { 2 n + 3 }$$

Evaluate

$$\frac{4n^{2}+8n+9}{3\left(2n+1\right)\left(2n+3\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 2n+1 is 3\left(2n+1\right). Multiply \frac{1}{3} times \frac{2n+1}{2n+1}. Multiply \frac{1}{2n+1} times \frac{3}{3}.

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

Since \frac{2n+1}{3\left(2n+1\right)} and \frac{3}{3\left(2n+1\right)} have the same denominator, add them by adding their numerators.

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

Combine like terms in 2n+1+3.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\left(2n+1\right) and 2n+3 is 3\left(2n+1\right)\left(2n+3\right). Multiply \frac{2n+4}{3\left(2n+1\right)} times \frac{2n+3}{2n+3}. Multiply \frac{1}{2n+3} times \frac{3\left(2n+1\right)}{3\left(2n+1\right)}.

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

Since \frac{\left(2n+4\right)\left(2n+3\right)}{3\left(2n+1\right)\left(2n+3\right)} and \frac{3\left(2n+1\right)}{3\left(2n+1\right)\left(2n+3\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(2n+4\right)\left(2n+3\right)-3\left(2n+1\right).

$$\frac{4n^{2}+6n+8n+12-6n-3}{3\left(2n+1\right)\left(2n+3\right)}$$

Combine like terms in 4n^{2}+6n+8n+12-6n-3.

$$\frac{4n^{2}+8n+9}{3\left(2n+1\right)\left(2n+3\right)}$$

Expand 3\left(2n+1\right)\left(2n+3\right).

$$\frac{4n^{2}+8n+9}{12n^{2}+24n+9}$$

Factor

$$\frac{4n^{2}+8n+9}{3\left(2n+1\right)\left(2n+3\right)}$$