Question Number 136519 by Dwaipayan Shikari last updated on 22/Mar/21
$$\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}\right)^{\mathrm{2}{k}} −\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}\right)^{\mathrm{2}{k}} +\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}}\right)^{\mathrm{2}{k}} −.... \\ $$$${Find}\:{the}\:{general}\:{form} \\ $$
Answered by mindispower last updated on 22/Mar/21
![1+Σ_(n≥1) (((−1)^n .Π_(m=0) ^(n−1) (2m+1)^(2k) )/(Π_(m=0) ^(n−1) .(2(m+1))^(2k) )) =1+Σ_(n≥1) .(−1)^n ((Π_(m=0) ^(n−1) (m+(1/2))^(2k) )/(Π_(m=0) ^(m=n−1) (m+1)^(2k) )) =1+Σ_(n≥1) (((Π_(m≤n−1) ((1/2)+m))^(2k) )/((Π_(m≤n−1) (1+m))^(2k−1) )).(((−1)^n )/(n!)) =1+Σ_(n≥1) .((((1/2))_n ^(2k) .)/((1)^(2k−1) )).(((−1)^n )/(n!)) =_(2k) F_(2k−1) determinant (((((1/2)),.........((1/2)))),((;(1),(1).......(1);[−1])))](Q136525.png)
$$\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} .\underset{{m}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}{k}} }{\underset{{m}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}.\left(\mathrm{2}\left({m}+\mathrm{1}\right)\right)^{\mathrm{2}{k}} } \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}.\left(−\mathrm{1}\right)^{{n}} \frac{\underset{{m}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({m}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}{k}} }{\underset{{m}=\mathrm{0}} {\overset{{m}={n}−\mathrm{1}} {\prod}}\left({m}+\mathrm{1}\right)^{\mathrm{2}{k}} } \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\underset{{m}\leqslant{n}−\mathrm{1}} {\prod}\left(\frac{\mathrm{1}}{\mathrm{2}}+{m}\right)\right)^{\mathrm{2}{k}} }{\left(\underset{{m}\leqslant{n}−\mathrm{1}} {\prod}\left(\mathrm{1}+{m}\right)\right)^{\mathrm{2}{k}−\mathrm{1}} }.\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!} \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}.\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}{k}} .}{\left(\mathrm{1}\right)^{\mathrm{2}{k}−\mathrm{1}} }.\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!} \\ $$$$=_{\mathrm{2}{k}} {F}_{\mathrm{2}{k}−\mathrm{1}} \begin{vmatrix}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right),.........\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\\{;\left(\mathrm{1}\right),\left(\mathrm{1}\right).......\left(\mathrm{1}\right);\left[−\mathrm{1}\right]}\end{vmatrix} \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 22/Mar/21
$${Thanks}\:{sir}\:!\: \\ $$
Commented by mindispower last updated on 23/Mar/21
$${pleasur} \\ $$