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The value of \(\cfrac{(1^4+\frac14)(3^4+\frac14)...((2n-1)^4+\frac14)}{(2^4+\frac14)(4^4+\frac14)....((2n)^4+\frac14)}\) is equal to

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Let P(n) = \(\cfrac{(1^4+\frac14)(3^4+\frac14)...((2n-1)^4+\frac14)}{(2^4+\frac14)(4^4+\frac14)....((2n)^4+\frac14)}\)

Put n = 1, we get P(1) = \(\cfrac{1^4+\frac14}{2^4+\frac14}=\frac{1+\frac14}{16+\frac14}\) = \(\cfrac{\frac54}{\frac{65}4}\)

\(\frac5{65}=\frac1{13}\)

From option (a), \(\frac{1}{4n^2+2n+1}\) at n = 1 = \(\frac1{4+2+1}=\frac17\)

Hence, (a) is wrong.

From option (b), \(\frac1{8n^2+4n+1}\) at n = 1 = \(\frac1{8 + 4 + 1}=\frac1{13}\)

Hence, option(b) may be true.

From option (c), \(\frac1{4(2n^2+n+1)}\) at n = 1 = \(\frac1{4(2+1+1)}=\frac1{16}\)

Hence, option (c) is wrong.

From option (d), \(\frac1{4(8n^2-4n+1)}\) at n = 1 = \(\frac{1}{8-4+1}=\frac15\)

Hence, option(d) is wrong.

Only option (b) may be true.

\(\therefore\) P(n) = \(\frac1{8n^2+4n+1}\) which can be proved by mathematical induction.

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