1. **State the problem:** We are given the function $$G_{q,\nu,r}[a,t] = \sum_{j=0}^\infty \frac{\Gamma(1-r)(-a)^j t^{(r+j)q-\nu-1}}{\Gamma(1+j)\Gamma(1-j-r)\Gamma((r+j)q-\nu)},$$ with conditions \(\text{Re}(qr-\nu)>0\), \(\text{Re}(s)>0\), and \(\left|\frac{a}{s^q}\right|<1\). 2. **Laplace transform given:** $$L \{ G_{q,\nu,r}[a,t] \} = \frac{s^\nu}{(s^q - a)^r},$$ with \(\text{Re}(qr-\nu)>0\), \(\text{Re}(s)>0\), and \(\left|\frac{a}{s^q}\right|>0\). 3. **Gamma function identities:** - For integer \(n\geq 0\), $$\Gamma(x-n) = \frac{\Gamma(x)}{(x-1)(x-2) \cdots (x-n)} = \frac{(-1)^n \Gamma(x)}{(1-x)_n},$$ where \((1-x)_n\) is the Pochhammer symbol. - Specifically, $$\Gamma(1-j-r) = \frac{\Gamma(1-r)}{(-r)(-1-r) \cdots (1-r-j)} = \frac{(-1)^j \Gamma(1-r)}{(r)_j},$$ for \(j=1,2,\ldots\). 4. **Rewrite the summand using the identity for \(\Gamma(1-j-r)\):** Substitute into the summand denominator: $$\Gamma(1-j-r) = \frac{(-1)^j \Gamma(1-r)}{(r)_j} \implies \frac{1}{\Gamma(1-j-r)} = \frac{(r)_j}{(-1)^j \Gamma(1-r)}.$$ 5. **Simplify the summand:** The summand is $$\frac{\Gamma(1-r)(-a)^j t^{(r+j)q-\nu-1}}{\Gamma(1+j) \Gamma(1-j-r) \Gamma((r+j)q-\nu)} = \frac{\Gamma(1-r)(-a)^j t^{(r+j)q-\nu-1}}{\Gamma(1+j) \Gamma(1-j-r) \Gamma((r+j)q-\nu)}.$$ Replace \(\frac{1}{\Gamma(1-j-r)}\) with the expression above: $$= \frac{\Gamma(1-r)(-a)^j t^{(r+j)q-\nu-1}}{\Gamma(1+j) \Gamma((r+j)q-\nu)} \cdot \frac{(r)_j}{(-1)^j \Gamma(1-r)} = \frac{(-a)^j t^{(r+j)q-\nu-1} (r)_j}{\Gamma(1+j) \Gamma((r+j)q-\nu) (-1)^j}.$$ 6. **Simplify the powers of \(-1\):** Note that \((-a)^j = (-1)^j a^j\), so $$\frac{(-a)^j}{(-1)^j} = a^j.$$ 7. **Final simplified summand:** $$\frac{a^j (r)_j t^{(r+j)q-\nu-1}}{\Gamma(1+j) \Gamma((r+j)q-\nu)}.$$ 8. **Interpretation:** The summation becomes $$G_{q,\nu,r}[a,t] = \sum_{j=0}^\infty \frac{a^j (r)_j t^{(r+j)q-\nu-1}}{\Gamma(1+j) \Gamma((r+j)q-\nu)}.$$ 9. **Summary:** We used the Gamma function identity to rewrite the summand in a simpler form involving the Pochhammer symbol \((r)_j\) and powers of \(a\) and \(t\). 10. **Graphical expression:** The numerator of the summand includes the product \((-r)(1-r) \cdots (1-j-r) (-a)^j t^{(r+j)q-\nu-1}\), which corresponds to the Pochhammer symbol \((r)_j\) times \(a^j t^{(r+j)q-\nu-1}\). **Final answer:** $$G_{q,\nu,r}[a,t] = \sum_{j=0}^\infty \frac{a^j (r)_j t^{(r+j)q-\nu-1}}{\Gamma(1+j) \Gamma((r+j)q-\nu)}.$$