samples/texts/1085779/page_7.md

where

$$\mathrm{\Omega }(a,i)=\frac{1}{i{4}^{m-i}(1-\frac{1}{a}{)}^i}\sum _{k=0}^{i-1}\frac{(1-\frac{1}{a}{)}^k}{k!}{\left(\frac{1}{2}\right)}_k\backslash Omega\; (a,\; i)\; =\; \backslash frac\{1\}\{i\; 4^\{m-i\}\; (1\; -\; \backslash frac\{1\}\{a\})^i\}\; \backslash sum\_\{k=0\}^\{i-1\}\; \backslash frac\{(1\; -\; \backslash frac\{1\}\{a\})^k\}\{k!\}\; \backslash left(\backslash frac\{1\}\{2\}\backslash right)\_k$$

we came to know that

$$I=\frac{(-1{)}^m}{m!}\frac{{\mathrm{\partial }}^m}{\mathrm{\partial }{a}^m}\left(\frac{{\tanh }^{-1}\left(\frac{1}{\sqrt{a}}\right)}{\sqrt{a}}\right)I\; =\; \backslash frac\{(-1)^m\}\{m!\}\; \backslash frac\{\backslash partial^m\}\{\backslash partial\; a^m\}\; \backslash left(\; \backslash frac\{\backslash tanh^\{-1\}\backslash left(\backslash frac\{1\}\{\backslash sqrt\{a\}\}\backslash right)\}\{\backslash sqrt\{a\}\}\; \backslash right)$$

∴

∴

$$H({\left(\frac{1}{a}\right)}_{r(m+1)};2_{r(m+1)};1)=a^{m+1}IH\backslash left(\backslash left(\backslash frac\{1\}\{a\}\backslash right)\_\{r(m+1)\};\; 2\_\{r(m+1)\};\; 1\backslash right)\; =\; a^\{m+1\}\; I$$

and finally we'll get the result

$$\begin{array}{cccc}& \mathcal{H}({\left(\frac{1}{a}\right)}_{r(m+1)};2_{r(m+1)};1)=\frac{1}{2}(\sum _{i=1}^m\left(\genfrac{}{}{0px}{}{2m-2i}{m-i}\right)\mathrm{\Omega }(a,i))+\frac{(2m)!}{2^{2m}(m!{)}^2{a}^{m+\frac{1}{2}}}{\tanh }^{-1}\left(\frac{1}{\sqrt{a}}\right)& & \text{(7)}\end{array}\backslash mathcal\{H\}\backslash left(\backslash left(\backslash frac\{1\}\{a\}\backslash right)\_\{r(m+1)\}\; ;\; 2\_\{r(m+1)\}\; ;\; 1\backslash right)\; =\; \backslash frac\{1\}\{2\}\; \backslash left(\; \backslash sum\_\{i=1\}^\{m\}\; \backslash binom\{2m-2i\}\{m-i\}\; \backslash Omega(a,i)\; \backslash right)\; +\; \backslash frac\{(2m)!\}\{2^\{2m\}(m!)^2\; a^\{m+\backslash frac\{1\}\{2\}\}\}\; \backslash tanh^\{-1\}\backslash left(\backslash frac\{1\}\{\backslash sqrt\{a\}\}\backslash right)\; \backslash tag\{7\}$$

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Example 1. let us plug the value's $a = 3$ and $m = 4$ in Eq(7) i.e $\mathcal{H}\left(\left(\frac{1}{3}\right){r(5)} ; 2{r(5)} ; 1\right)$ where 3 and 2 are repeated 5 times. Therefore from theorem 1 we'll get