Solve 1+1/2(1/2n-1-1/2n+1)

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

Evaluate

$$\frac{4n^{2}}{4n^{2}-1}$$

Short Solution Steps

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

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

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

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

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

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

Combine like terms in 2n+1-2n+1.

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

Multiply \frac{1}{2} times \frac{2}{\left(2n-1\right)\left(2n+1\right)} by multiplying numerator times numerator and denominator times denominator.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2\left(2n-1\right)\left(2n+1\right)}{2\left(2n-1\right)\left(2n+1\right)}.

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

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

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

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

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

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

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

Cancel out 2 in both numerator and denominator.

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

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

$$\frac{4n^{2}}{4n^{2}-1}$$

Expand

$$\frac{4n^{2}}{4n^{2}-1}$$

Short Solution Steps

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

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

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

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

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

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

Combine like terms in 2n+1-2n+1.

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

Multiply \frac{1}{2} times \frac{2}{\left(2n-1\right)\left(2n+1\right)} by multiplying numerator times numerator and denominator times denominator.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2\left(2n-1\right)\left(2n+1\right)}{2\left(2n-1\right)\left(2n+1\right)}.

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

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

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

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

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

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

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

Cancel out 2 in both numerator and denominator.

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

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

$$\frac{4n^{2}}{4n^{2}-1}$$