∏n=113(1−4(2n−1)2)=?\prod_{n=1}^{13}\left(1-\frac{4}{(2n-1)^2}\right)=?n=1∏13(1−(2n−1)24)=?
asked by guest on Jan 25, 2025 at 11:13 am
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MathBot Answer:
∏n=1131−4(2n−1)2=−2725\displaystyle \prod_{n = 1}^{13}{1 - \frac{4}{{\left( 2 n - 1 \right)}^{2}}} = - \frac{27}{25} n=1∏131−(2n−1)24=−2527
∏n=1131−4(2n−1)2=∏n=113(1−44n2−4n+1)\prod_{n = 1}^{13}{1 - \frac{4}{{\left( 2 n - 1 \right)}^{2}}} = \prod_{n=1}^{13} \left(1 - \frac{4}{4 n^{2} - 4 n + 1}\right)n=1∏131−(2n−1)24=n=1∏13(1−4n2−4n+14)
