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


Is there any formulas I can use?



Sum of the first m terms of the expansion $(x+y)^n$
One can construct the relation using the identity
$$
\binom{n}{i} = \binom{n-1}{i} + \binom{n-1}{i-1}
$$
Then, writing the $i=0$ term separately,
\begin{eqnarray}
S(x,y,m,n) &=& y^n + \sum\_{i=1}^{m} \binom{n-1}{i} x^i y^{n-i}
+ \sum\_{i=1}^{m} \binom{n-1}{i-1} x^i y^{n-i} \\
&=& y^n + y\sum\_{i=1}^{m} \binom{n-1}{i} x^i y^{n-1-i}
+ \sum\_{i=0}^{m-1} \binom{n-1}{i} x^{i+1} y^{n-i-1} \\
&=& y S(x,y,m,n-1) + x\left[ S(x,y,m,n-1) - \binom{n-1}{m} x^m y^{n-1-m} \right] \\
&=& (x+y) S(x,y,m,n-1) \ - \ \binom{n-1}{m} x^{m+1} y^{n-1-m}
\end{eqnarray}


