Question

(1-1/2^2) (1-1/3^2) (1-1/4^2) ... (1-1/1000^2) = ?

Solution

The question can be written in this way as well

{1- (1/2)^2}{1- (1/3)^2}...

= {1+(1/2)} {1-(1/2)}.... and so on ......using (a^2 - b^2) formula

= (3/2) (1/2) * (4/3) (2/3) * (5/4) (3/4) .... and so on

if we continue this way we see a sequence { (n+1) / n }* { (n-1) / n } = { (n^2 - 1) / n^2)

if we find the product of the first four terms we see the result is { (3/4) * (8/9) * (15/16) * (24/25) }= 3/5

then the product of the first eight term is 5/9, then the product of the first twelve term is 7/13.. then the product of the first sixteen term is 9/15 and so on....

So product of first number of terms, n is [ { (n/2)+1 } / (n + 1) ]

So product of first 1000th term is 501/1001 ..

Well actually the solution is easy it depends on how fast it clicks on your mind...

(1-1/2^2) (1-1/3^2) (1-1/4^2) ... (1-1/1000^2) = ?

Solution

The question can be written in this way as well

{1- (1/2)^2}{1- (1/3)^2}...

= {1+(1/2)} {1-(1/2)}.... and so on ......using (a^2 - b^2) formula

= (3/2) (1/2) * (4/3) (2/3) * (5/4) (3/4) .... and so on

if we continue this way we see a sequence { (n+1) / n }* { (n-1) / n } = { (n^2 - 1) / n^2)

if we find the product of the first four terms we see the result is { (3/4) * (8/9) * (15/16) * (24/25) }= 3/5

then the product of the first eight term is 5/9, then the product of the first twelve term is 7/13.. then the product of the first sixteen term is 9/15 and so on....

So product of first number of terms, n is [ { (n/2)+1 } / (n + 1) ]

So product of first 1000th term is 501/1001 ..

Well actually the solution is easy it depends on how fast it clicks on your mind...