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Let $S(x, y, m_1,m_2, n) = \sum\limits_{i=m_1}^{m_2} \binom{n}{i}x^i y^{n-i}$, where $0 < m_1\leq m_2 < n$. I want to derive the relation between $S(x, y, m_1, m_2, n)$ and $S(x, y, m_1-1, m_2, n-1)$.

Is there any formulas I can use? Similarly to Sum of the first m terms of the expansion $(x+y)^n$.

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Use the fact that, with $$ T(x, y, m, n) = \sum\limits_{i=0}^m \binom{n}{i}x^i y^{n-i} $$ you have $$ S(x, y, m_1, m_2, n)= T(x, y, m_2, n)-T(x, y, m_1-1, n) $$

Now using the formula for $T$ in the provided link, you can connect now $S(x,y,m_1, m_2,n)$ to $S(x,y,m_1, m_2,n-1)$. Finally $$ S(x,y,m_1, m_2,n-1)+\binom{n}{m_1-1}x^{m_1-1} y^{n-m_1}=S(x,y,m_1-1, m_2,n-1) $$ allows you connect the two you want.

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  • $\begingroup$ You refer to "the provided link", but you do not seem to have provided one. $\endgroup$
    – LSpice
    Commented Jan 20, 2022 at 4:22
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    $\begingroup$ @LSpice The OP states this problem as follow-up to a question. And he provided the link. $\endgroup$
    – Nick S
    Commented Jan 20, 2022 at 5:31
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    $\begingroup$ Ah, then the provided link is Sum of the first m terms of the expansion $(x+y)^n$. $\endgroup$
    – LSpice
    Commented Nov 16, 2022 at 19:43

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