I need to calculate this summation: $$\sum_{i=j}^m {m\choose i} r^{i}$$ I know $\sum_{i=0}^m {m\choose i} r^{i}=(1+r)^m$, but is there a nice closed form answer for the summation above?
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1There is a closed form for $r = -1$, but I do not believe that it is known for other values of $r$ (except for the obvious cases $j=0$ and $j=m$). – Siméon Jan 08 '14 at 16:15
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2There's no closed form for it, but here are two links that can help: http://math.stackexchange.com/questions/69532/partial-sum-of-rows-of-pascals-triangle http://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n/17203#17203 – Alex Jan 08 '14 at 16:39
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The links are helpful, thanks! – Mah Jan 08 '14 at 16:42
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$$\sum_{i=0}^{m}\binom{m}{i}r^{i}=\sum_{i=0}^{j-1}\binom{m}{i}r^{i}+\sum_{i=j}^m \binom{m}{i}r^{i}=(1+r)^m$$ $$\sum_{i=j}^m \binom{m}{i}r^{i}=(1+r)^m-\sum_{i=0}^{j-1}\binom{m}{i}r^{i}$$
Adi Dani
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4I don't think that's what is being asked. I agree it's true but it's not really a closed form and doesn't simplify the expression much. – user88595 Jan 08 '14 at 16:11
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I am not convinced that there exists such simple formula. Above formula make sense if $j<m/2$ – Adi Dani Jan 08 '14 at 16:15
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It's basically the last $m-j+1$ terms of the expansion of $(1+r)^m$, and I don't really think there's a nice closed form of that that just depends on $r$, $m$, and $j$.
Nishant
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