where

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

we came to know that

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

∴

$$
\begin{align*}
I &= \frac{(-1)^m}{m!} \left( \frac{(-1)^m m!}{2a^{m+1}} \left( \sum_{i=1}^{m} \binom{2m-2i}{m-i} \Omega(a,i) \right) + \frac{(-1)^m (2m)!}{2^{2m}(m!)a^{m+\frac{1}{2}}} \tanh^{-1}\left(\frac{1}{\sqrt{a}}\right) \right) \\
&= \frac{1}{2a^{m+1}} \left( \sum_{i=1}^{m} \binom{2m-2i}{m-i} \Omega(a,i) \right) + \frac{(2m)!}{2^{2m}(m!)^2 a^{m+\frac{1}{2}}} \tanh^{-1}\left(\frac{1}{\sqrt{a}}\right)
\end{align*}
$$

∴

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

and finally we'll get the result

$$
\mathcal{H}\left(\left(\frac{1}{a}\right)_{r(m+1)} ; 2_{r(m+1)} ; 1\right) = \frac{1}{2} \left( \sum_{i=1}^{m} \binom{2m-2i}{m-i} \Omega(a,i) \right) + \frac{(2m)!}{2^{2m}(m!)^2 a^{m+\frac{1}{2}}} \tanh^{-1}\left(\frac{1}{\sqrt{a}}\right) \tag{7}
$$

□

**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

$$
\begin{align*}
\mathcal{H}\left(\left(\frac{1}{3}\right)_{r(5)} ; 2_{r(5)} ; 1\right) &= \frac{1}{2} \left( \sum_{i=1}^{4} \binom{8-2i}{4-i} \Omega(3,i) \right) + \frac{(8)!\sqrt{3}}{2^8 (4!)^2} \tanh^{-1}\left(\frac{1}{\sqrt{3}}\right) \\
&= \frac{249}{128} + \frac{35\sqrt{3}}{128} \tanh^{-1}\left(\frac{1}{\sqrt{3}}\right)
\end{align*}
$$