Solve (1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)*L*(1+frac{1}{99})*(1-frac{1}{99})

$$( 1 + \frac { 1 } { 2 } ) \times ( 1 - \frac { 1 } { 2 } ) \times ( 1 + \frac { 1 } { 3 } ) \times ( 1 - \frac { 1 } { 3 } ) \times L \times ( 1 + \frac { 1 } { 99 } ) \times ( 1 - \frac { 1 } { 99 } )$$

Evaluate

$$\frac{19600L}{29403}$$

Solution Steps

Convert 1 to fraction \frac{2}{2}.

$$\left(\frac{2}{2}+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{2+1}{2}\left(1-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Add 2 and 1 to get 3.

$$\frac{3}{2}\left(1-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{2}{2}.

$$\frac{3}{2}\left(\frac{2}{2}-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{3}{2}\times \frac{2-1}{2}\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Subtract 1 from 2 to get 1.

$$\frac{3}{2}\times \frac{1}{2}\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{3\times 1}{2\times 2}\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Do the multiplications in the fraction \frac{3\times 1}{2\times 2}.

$$\frac{3}{4}\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{3}{4}\left(\frac{3}{3}+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{3}{4}\times \frac{3+1}{3}\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Add 3 and 1 to get 4.

$$\frac{3}{4}\times \frac{4}{3}\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Cancel out \frac{3}{4} and its reciprocal \frac{4}{3}.

$$\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{3}{3}.

$$\left(\frac{3}{3}-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{3-1}{3}L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Subtract 1 from 3 to get 2.

$$\frac{2}{3}L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{99}{99}.

$$\frac{2}{3}L\left(\frac{99}{99}+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Since \frac{99}{99} and \frac{1}{99} have the same denominator, add them by adding their numerators.

$$\frac{2}{3}L\times \frac{99+1}{99}\left(1-\frac{1}{99}\right)$$

Add 99 and 1 to get 100.

$$\frac{2}{3}L\times \frac{100}{99}\left(1-\frac{1}{99}\right)$$

Multiply \frac{2}{3} times \frac{100}{99} by multiplying numerator times numerator and denominator times denominator.

$$\frac{2\times 100}{3\times 99}L\left(1-\frac{1}{99}\right)$$

Do the multiplications in the fraction \frac{2\times 100}{3\times 99}.

$$\frac{200}{297}L\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{99}{99}.

$$\frac{200}{297}L\left(\frac{99}{99}-\frac{1}{99}\right)$$

Since \frac{99}{99} and \frac{1}{99} have the same denominator, subtract them by subtracting their numerators.

$$\frac{200}{297}L\times \frac{99-1}{99}$$

Subtract 1 from 99 to get 98.

$$\frac{200}{297}L\times \frac{98}{99}$$

Multiply \frac{200}{297} times \frac{98}{99} by multiplying numerator times numerator and denominator times denominator.

$$\frac{200\times 98}{297\times 99}L$$

Do the multiplications in the fraction \frac{200\times 98}{297\times 99}.

$$\frac{19600}{29403}L$$

Expand

$$\frac{19600L}{29403}$$

Solution Steps

Convert 1 to fraction \frac{2}{2}.

$$\left(\frac{2}{2}+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{2+1}{2}\left(1-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Add 2 and 1 to get 3.

$$\frac{3}{2}\left(1-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{2}{2}.

$$\frac{3}{2}\left(\frac{2}{2}-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{3}{2}\times \frac{2-1}{2}\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Subtract 1 from 2 to get 1.

$$\frac{3}{2}\times \frac{1}{2}\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{3\times 1}{2\times 2}\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Do the multiplications in the fraction \frac{3\times 1}{2\times 2}.

$$\frac{3}{4}\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{3}{4}\left(\frac{3}{3}+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{3}{4}\times \frac{3+1}{3}\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Add 3 and 1 to get 4.

$$\frac{3}{4}\times \frac{4}{3}\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Cancel out \frac{3}{4} and its reciprocal \frac{4}{3}.

$$\left(1-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{3}{3}.

$$\left(\frac{3}{3}-\frac{1}{3}\right)L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

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

$$\frac{3-1}{3}L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Subtract 1 from 3 to get 2.

$$\frac{2}{3}L\left(1+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{99}{99}.

$$\frac{2}{3}L\left(\frac{99}{99}+\frac{1}{99}\right)\left(1-\frac{1}{99}\right)$$

Since \frac{99}{99} and \frac{1}{99} have the same denominator, add them by adding their numerators.

$$\frac{2}{3}L\times \frac{99+1}{99}\left(1-\frac{1}{99}\right)$$

Add 99 and 1 to get 100.

$$\frac{2}{3}L\times \frac{100}{99}\left(1-\frac{1}{99}\right)$$

Multiply \frac{2}{3} times \frac{100}{99} by multiplying numerator times numerator and denominator times denominator.

$$\frac{2\times 100}{3\times 99}L\left(1-\frac{1}{99}\right)$$

Do the multiplications in the fraction \frac{2\times 100}{3\times 99}.

$$\frac{200}{297}L\left(1-\frac{1}{99}\right)$$

Convert 1 to fraction \frac{99}{99}.

$$\frac{200}{297}L\left(\frac{99}{99}-\frac{1}{99}\right)$$

Since \frac{99}{99} and \frac{1}{99} have the same denominator, subtract them by subtracting their numerators.

$$\frac{200}{297}L\times \frac{99-1}{99}$$

Subtract 1 from 99 to get 98.

$$\frac{200}{297}L\times \frac{98}{99}$$

Multiply \frac{200}{297} times \frac{98}{99} by multiplying numerator times numerator and denominator times denominator.

$$\frac{200\times 98}{297\times 99}L$$

Do the multiplications in the fraction \frac{200\times 98}{297\times 99}.

$$\frac{19600}{29403}L$$