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Given $$\sum\limits_{i=0}^\infty a_i z^i=A(z)$$ and $$\sum\limits_{i=1}^\infty b_i z^i=B(z)$$ and $$\sum\limits_{i=0}^\infty c_i z^i=C(z)$$ Find $\sum\limits_{i=1}^\infty\sum\limits_{j=0}^\infty a_j \sum\limits_{k=1}^\infty b_kc_{k+j-i} z^i$ in terms of the generating functions $A(z),B(z),C(z)$.

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Since $i=(j)+(k)-(j+k-i)$, the triple sum is $$\sum\limits_{\alpha,\beta,\gamma}a_\alpha b_\beta c_\gamma z^{\alpha+\beta-\gamma}=A(z)B(z)C(1/z).$$

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Thank you for your help. Please see this also $\sum\limits_{i=0}^\infty a_i z^i=A(z)$ and $\sum\limits_{j=1}^\infty b_j z^j=B(z)$ and $\sum\limits_{k=0}^\infty c_k z^{-k}=C(1/z)$ Then $\sum\limits_{i=0}^\infty\sum\limits_{j=1}^\infty\sum\limits_{k=1}^\infty a_ib_jc_{i+j-k} z^k=$? – Litun John Feb 20 '13 at 6:40
What are you asking me to do? – Did Feb 20 '13 at 6:53
Whether this expression is same as the previous one $A(z)B(z)C(1/z)$ – Litun John Feb 20 '13 at 8:13
As $A(z)B(z)=\sum\limits_{i=0}^{\infty}\sum\limits_{j=0}^{i}a_jb_{i-j}$ then what is the expression for $A(z)B(z)C(1/z)$ – Litun John Feb 20 '13 at 8:27

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