The Progression is:
$0(n-0) + 1(n-1) + 3(n-2) + 6(n-3) + 10(n-4) + .... $
which can be represented as:
$\sum_{i = 0}^n \frac{i(i+1)}{2}(n-i)$
Is there a general formula for this summation?
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1$\sum_{i = 0}^n \frac{i(i+1)}{2}(n-i)=\frac{1}{24}n(n-1)(n+1)(n+2)$ by WolframAlpha. – Yuta May 09 '19 at 12:13
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How do you arrive at that? – srt1104 May 09 '19 at 12:15
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Did you mean for the $n$ in the upper bound of summation to be the same as the $n$ in the summand? If not, it's an abuse of notation. – Deepak May 09 '19 at 12:15
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yes, they are the same. – srt1104 May 09 '19 at 12:16
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Possible duplicate of Methods to compute $\sum_{k=1}^nk^p$ without Faulhaber's formula – Simply Beautiful Art May 09 '19 at 12:47
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1To neatly get @Yuta's result from WolframAlpha, you can apply summation by parts twice with the hockey stick identity. – Simply Beautiful Art May 09 '19 at 12:48
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Also, your identity is a special case of https://math.stackexchange.com/q/1938753/177399 – Mike Earnest May 09 '19 at 20:32
2 Answers
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\begin{align} \sum_i \frac{i(i+1)}2(n-i) &=\sum_i \binom{i+1}2(n+2-(i+2)) \\&=(n+2)\sum_i \binom{i+1}2-\sum_i (i+2)\binom{i+1}2 \\&=(n+2)\sum_i \binom{i+1}2-3\sum_i\binom{i+2}3 \\&\stackrel{\text{H.S.}}=(n+2)\binom{n+2}3-3\binom{n+3}4 \\&=(n+3)\binom{n+2}3-\binom{n+2}3-3\binom{n+3}4 \\&=4\binom{n+3}4-\binom{n+2}3-3\binom{n+3}4 \\&=\binom{n+3}4-\binom{n+2}3 \\&= \binom{n+2}4 \end{align} $\stackrel{\text{H.S.}}=$ is the hockey stick identity.
Mike Earnest
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$\sum_{i = 0}^n \frac{i(i+1)}{2}(n-i)=\sum_{i = 0}^n \frac1 {2}(i^2n-i^3+in-i^2)=(n-1)\sum i^2-\sum i^3+n\sum i $. Now, $\sum i=\frac {n(n+1)} {2} ; \sum i^2=\frac {n(n+1)(2n+1)} 6;\sum i^3= \frac {n^2(n+1)^2} 4$
Tojra
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