1thanks sirji
find the product
\frac{(1^4 +\frac{1}{4})(3^4 +\frac{1}{4})(5^4 +\frac{1}{4}).......(49^4 +\frac{1}{4})}{(2^4 +\frac{1}{4})(4^4 +\frac{1}{4})(6^4 +\frac{1}{4}).......(50^4 +\frac{1}{4})}
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3 Answers
1\prod_{r=1}^{25}{}\frac{4(2r-1)^4+1}{4(2r)^4+1}
=\prod_{r=1}^{25}{}\frac{(2(2r-1)^2)^2+1}{(2(2r)^2)^2+1}
=\prod_{r=1}^{25}{}\frac{(2(2r-1)^2+1)^2-2(2(2r-1)^2)(1)}{(2(2r)^2+1)^2-2(2(2r)^2)(1)}
=\prod_{r=1}^{25}{}\frac{(8r^2-8r+3)^2-(4r-2)^2}{(8r^2+1)^2-(4r)^2}
=\prod_{r=1}^{25}{}\frac{(8r^2-12r+5)(8r^2-4r+1)}{(8r^2-4r+1)(8r^2+4r+1)}
=\prod_{r=1}^{25}{}\frac{(8r^2-12r+5)}{(8r^2+4r+1)}
= 113 1341 .......
= 18(252)+4(25)+1