Equal to what?

Algebra Level 2

( 1 − 1 y ) ( 1 − 1 y + 1 ) ( 1 − 1 y + 2 ) ⋯ ( 1 − 1 y + y ) = ? \large \left(1 - \frac{1}{y}\right)\left(1 - \frac{1}{y+1}\right) \left(1 - \frac{1}{y+2}\right) \cdots \left(1 - \frac{1}{y+y}\right)= \ ?

1 2 y \frac{1}{2y} 2 y y − 1 \frac{2y}{y-1} 1 y \frac{1}{y} y − 1 2 y \frac{y-1}{2y}

Ojasee Duble by Ojasee Duble , 19, India

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1 solution

= ( 1 − 1 y ) ( 1 − 1 y + 1 ) ( 1 − 1 y + 2 ) . . . ( 1 − 1 y + y ) = y − 1 y ⋅ y + 1 − 1 y + 1 ⋅ y + 2 − 1 y + 2 ⋅ . . . ⋅ y + y − 1 y + y = y − 1 y ⋅ y y + 1 ⋅ y + 1 y + 2 ⋅ . . . ⋅ 2 y − 1 2 y = y − 1 2 y \large =\left( 1-\frac { 1 }{ y } \right) \left( 1-\frac { 1 }{ y+1 } \right) \left( 1-\frac { 1 }{ y+2 } \right) ...\left( 1-\frac { 1 }{ y+y } \right) \\ \large =\frac { y-1 }{ y } \cdot \frac { y+1-1 }{ y+1 } \cdot \frac { y+2-1 }{ y+2 } \cdot ...\cdot \frac { y+y-1 }{ y+y } \\ \large =\frac { y-1 }{ y } \cdot \frac { y }{ y+1 } \cdot \frac { y+1 }{ y+2 } \cdot ...\cdot \frac { 2y-1 }{ 2y } \\ \large=\frac { y-1 }{ 2y }

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How did you get the 4th line of your solution?

Ojasee Duble - 3 years, 9 months ago

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See the numerator and the denominator looking at the pattern.

Wildan Bagus Wicaksono - 3 years, 9 months ago

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Oh yeah...Ok...Got it!! Thanks :)

Ojasee Duble - 3 years, 9 months ago

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