Solve 1/[frac{1{3}+2/3(n-1)](1/3+2n/3)}

$$\frac { 1 } { [ \frac { 1 } { 3 } + \frac { 2 } { 3 } ( n - 1 ) ] ( \frac { 1 } { 3 } + \frac { 2 n } { 3 } ) }$$

Evaluate

$$\frac{9}{4n^{2}-1}$$

Short Solution Steps

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

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

Use the distributive property to multiply \frac{2}{3} by n-1.

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

Multiply \frac{2}{3} and -1 to get -\frac{2}{3}.

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

Since \frac{1}{3} and \frac{2}{3} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 2 from 1 to get -1.

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

Use the distributive property to multiply -\frac{1}{3}+\frac{2}{3}n by \frac{1+2n}{3}.

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

Multiply -\frac{1}{3} times \frac{1+2n}{3} by multiplying numerator times numerator and denominator times denominator.

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

Multiply \frac{2}{3} times \frac{1+2n}{3} by multiplying numerator times numerator and denominator times denominator.

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

Multiply 3 and 3 to get 9.

$$\frac{1}{\frac{-\left(1+2n\right)}{9}+\frac{2\left(1+2n\right)}{3\times 3}n}$$

Multiply 3 and 3 to get 9.

$$\frac{1}{\frac{-\left(1+2n\right)}{9}+\frac{2\left(1+2n\right)}{9}n}$$

Express \frac{2\left(1+2n\right)}{9}n as a single fraction.

$$\frac{1}{\frac{-\left(1+2n\right)}{9}+\frac{2\left(1+2n\right)n}{9}}$$

Since \frac{-\left(1+2n\right)}{9} and \frac{2\left(1+2n\right)n}{9} have the same denominator, add them by adding their numerators.

$$\frac{1}{\frac{-\left(1+2n\right)+2\left(1+2n\right)n}{9}}$$

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

$$\frac{1}{\frac{-1-2n+2n+4n^{2}}{9}}$$

Combine like terms in -1-2n+2n+4n^{2}.

$$\frac{1}{\frac{-1+4n^{2}}{9}}$$

Divide 1 by \frac{-1+4n^{2}}{9} by multiplying 1 by the reciprocal of \frac{-1+4n^{2}}{9}.

$$\frac{9}{-1+4n^{2}}$$

Expand

$$\frac{9}{4n^{2}-1}$$

Short Solution Steps

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

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

Use the distributive property to multiply \frac{2}{3} by n-1.

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

Multiply \frac{2}{3} and -1 to get -\frac{2}{3}.

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

Since \frac{1}{3} and \frac{2}{3} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 2 from 1 to get -1.

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

Use the distributive property to multiply -\frac{1}{3}+\frac{2}{3}n by \frac{1+2n}{3}.

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

Multiply -\frac{1}{3} times \frac{1+2n}{3} by multiplying numerator times numerator and denominator times denominator.

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

Multiply \frac{2}{3} times \frac{1+2n}{3} by multiplying numerator times numerator and denominator times denominator.

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

Multiply 3 and 3 to get 9.

$$\frac{1}{\frac{-\left(1+2n\right)}{9}+\frac{2\left(1+2n\right)}{3\times 3}n}$$

Multiply 3 and 3 to get 9.

$$\frac{1}{\frac{-\left(1+2n\right)}{9}+\frac{2\left(1+2n\right)}{9}n}$$

Express \frac{2\left(1+2n\right)}{9}n as a single fraction.

$$\frac{1}{\frac{-\left(1+2n\right)}{9}+\frac{2\left(1+2n\right)n}{9}}$$

Since \frac{-\left(1+2n\right)}{9} and \frac{2\left(1+2n\right)n}{9} have the same denominator, add them by adding their numerators.

$$\frac{1}{\frac{-\left(1+2n\right)+2\left(1+2n\right)n}{9}}$$

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

$$\frac{1}{\frac{-1-2n+2n+4n^{2}}{9}}$$

Combine like terms in -1-2n+2n+4n^{2}.

$$\frac{1}{\frac{-1+4n^{2}}{9}}$$

Divide 1 by \frac{-1+4n^{2}}{9} by multiplying 1 by the reciprocal of \frac{-1+4n^{2}}{9}.

$$\frac{9}{-1+4n^{2}}$$