Solve k/2k+1+1/(2k+1)(2k+3)

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

Evaluate

$$\frac{k+1}{2k+3}$$

Short Solution Steps

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

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

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

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

Do the multiplications in k\left(2k+3\right)+1.

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

Factor the expressions that are not already factored in \frac{2k^{2}+3k+1}{\left(2k+1\right)\left(2k+3\right)}.

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

Cancel out 2k+1 in both numerator and denominator.

$$\frac{k+1}{2k+3}$$

Expand

$$\frac{k+1}{2k+3}$$

Short Solution Steps

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

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

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

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

Do the multiplications in k\left(2k+3\right)+1.

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

Factor the expressions that are not already factored in \frac{2k^{2}+3k+1}{\left(2k+1\right)\left(2k+3\right)}.

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

Cancel out 2k+1 in both numerator and denominator.

$$\frac{k+1}{2k+3}$$