The product of #(1-1/(2^2))(1-1/(3^2))....(1-1/(9^2))(1-1/(10^2))=a/b# then find a and b without full simplification?

Question

1296 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-17T16:31:49

The answer is #a=11# and #b=20#

I think it's the same as

#Pi_(k=2)^10(1-1/k^2)#

But I did it this way

#(1-1/2^2)(1-1/3^2)(1-1/4^2)....(1-1/9^2)(1-1/10^2)#

#=3/4xx8/9xx15/16xx24/25xx35/36xx48/49xx63/64xx80/81xx99/100#

#=11/20#

The answer is #a=11# and #b=20#

Answer 2 · 2018-07-17T19:45:32

#(1-1/2^2)(1-1/3^2)(1-1/4^2)....(1-1/9^2)(1-1/10^2)#

#=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)....(1-1/9)(1+1/9)(1-1/10)(1+1/10)#

#1/2*3/2*2/3*4/3*3/4*5/4.....8/9*10/9*9/10*11/10#

#=1/2*11/10=11/20#

So #a=11andb=20#

The product of #(1-1/(2^2))*(1-1/(3^2))....(1-1/(9^2))*(1-1/(10^2))=a/b# then find a and b without full simplification?