Can you generalise this?

Algebra Level 4

∏ j = 2 1999 ( 1 − 1 j 2 ) \large \prod_{j=2}^{1999} \left( 1 - \dfrac1{j^2} \right)

If the value of the product above is equal to a b \dfrac ab , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 2999.

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Venkata Karthik Bandaru by Venkata Karthik Bandaru , 20, India

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1 solution

Vishal S
Dec 30, 2015

∏ j = 2 1999 ( 1 − 1 j 2 ) \prod _{ j=2 }^{ 1999 } \left( 1-\frac { 1 }{ { j }^{ 2 } } \right )

We can express ( 1 − 1 j 2 ) \left( 1-\frac { 1 }{ { j }^{ 2 } } \right) as ( 1 2 − ( 1 j ) 2 ) \left( { 1 }^{ 2 }-{ \left( \frac { 1 }{ j } \right) }^{ 2 } \right) = ( 1 − 1 j ) \left( 1-\frac { 1 }{ { j } } \right) ( 1 + 1 j ) \left( 1+\frac { 1 }{ { j } } \right)

⟹ \Longrightarrow ∏ j = 2 1999 ( 1 − 1 j 2 ) \prod _{ j=2 }^{ 1999 } \left( 1-\frac { 1 }{ { j }^{ 2 } } \right ) = ∏ j = 2 1999 ( 1 + 1 j ) ( 1 − 1 j ) =\prod _{ j=2 }^{ 1999 }\left( 1+\frac { 1 }{ j } \right) \left( 1-\frac { 1 }{ j } \right)

= [ ( 1 + 1 2 ) ( 1 + 1 3 ) ( 1 + 1 4 ) . . . . ( 1 + 1 1999 ) ] [ ( 1 − 1 2 ) ( 1 − 1 3 ) ( 1 − 1 4 ) . . . . ( 1 − 1 1999 ) ] =\left[ \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) \left( 1+\frac { 1 }{ 4 } \right) ....\left( 1+\frac { 1 }{ 1999 } \right) \right] \left[ \left( 1-\frac { 1 }{ 2 } \right) \left( 1-\frac { 1 }{ 3 } \right) \left( 1-\frac { 1 }{ 4 } \right) ....\left( 1-\frac { 1 }{ 1999 } \right) \right]

= [ ( 3 2 ) ( 4 3 ) ( 5 4 ) . . . . ( 2000 1999 ) ] [ ( 1 2 ) ( 2 3 ) ( 3 4 ) . . . . ( 1998 1999 ) ] =\left[ \left( \frac { 3 }{ 2 } \right) \left( \frac { 4 }{ 3 } \right) \left( \frac { 5 }{ 4 } \right) ....\left( \frac { 2000 }{ 1999 } \right) \right] \left[ \left( \frac { 1 }{ 2 } \right) \left( \frac { 2 }{ 3 } \right) \left( \frac { 3 }{ 4 } \right) ....\left( \frac { 1998 }{ 1999 } \right) \right]

= ( 2000 2 ) ( 1 1999 ) =\left( \frac { 2000 }{ 2 } \right) \left( \frac { 1 }{ 1999 } \right)

= 1000 1999 =\frac { 1000 }{ 1999 }

By comparing the above fraction with a b \frac { a }{ b } , we get

a=1000 & b=1999

∴ \therefore a+b=2999

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Can you mention the generalised identity ? By the way, nice solution :).

Venkata Karthik Bandaru - 5 years, 5 months ago

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Thx.I have mentioned the identity used to genralise

Vishal S - 5 years, 5 months ago

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(-_-') I guess you didnt get me. I meant a generalised simplification of the product.

Venkata Karthik Bandaru - 5 years, 5 months ago

Simplify ∏ j = 2 n ( 1 − 1 j 2 ) \large \prod_{j=2}^{n} \left( 1 - \dfrac1{j^2} \right) .

Venkata Karthik Bandaru - 5 years, 5 months ago

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@Venkata Karthik Bandaru – After simplification which I have done in my solution, the above product can be written as

[ ( 1 + 1 2 ) ( 1 + 1 3 ) ( 1 + 1 4 ) . . . . ( 1 + 1 n ) ] [ ( 1 − 1 2 ) ( 1 − 1 3 ) ( 1 − 1 4 ) . . . . ( 1 − 1 n ) ] \left[ \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) \left( 1+\frac { 1 }{ 4 } \right) ....\left( 1+\frac { 1 }{ n } \right) \right] \left[ \left( 1-\frac { 1 }{ 2 } \right) \left( 1-\frac { 1 }{ 3 } \right) \left( 1-\frac { 1 }{ 4 } \right) ....\left( 1-\frac { 1 }{ n } \right) \right]

= [ ( 3 2 ) ( 4 3 ) ( 5 4 ) . . . . ( n + 1 n ) ] [ ( 1 2 ) ( 2 3 ) ( 3 4 ) . . . . ( n − 1 n ) ] =\left[ \left( \frac { 3 }{ 2 } \right) \left( \frac { 4 }{ 3 } \right) \left( \frac { 5 }{ 4 } \right) ....\left( \frac { n+1 }{ n } \right) \right] \left[ \left( \frac { 1 }{ 2 } \right) \left( \frac { 2 }{ 3 } \right) \left( \frac { 3 }{ 4 } \right) ....\left( \frac { n-1 }{ n } \right) \right]

After cancellation, the resultant of the above product will be ( n + 1 2 ) ( 1 n ) \left( \frac { n+1 }{ 2 } \right) \left( \frac { 1 }{ n } \right)

= n + 1 2 n =\frac { n+1 }{ 2n }

Vishal S - 5 years, 5 months ago

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@Vishal S – That was what I asked you to mention :). Anyways, Happy New Year bro.

Venkata Karthik Bandaru - 5 years, 5 months ago

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@Venkata Karthik Bandaru – Happy new year

Vishal S - 5 years, 5 months ago

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