Are there any identities that deals with a summation like $\sum_{i=0}^{9} {12-i\choose3}{3+i\choose3}$.
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i got 11440 as the searched result – Dr. Sonnhard Graubner Oct 06 '15 at 17:39
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I want to find a way to evaluate that without a calculator. – 1-___- Oct 06 '15 at 17:44
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1Hint: Notice that the general term is a simple polynomial in i, and then use Faulhaber's formulas. – Lucian Oct 06 '15 at 19:43
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Take a $i$ in $\{0, ..., 9\}$. You can say : $${12 - i \choose 3}{3 + i \choose 3} = \frac{(12-i)!}{3!(9-i)!}\frac{(3+i)!}{3!i!} = \frac{1}{(3!)^2} \frac{(12-i)(11-i)(...)(2)(1)}{(9-i)(8-i)(...)(2)(1)}\frac{(3+i)(2+i)(1+i)(i)(...)(2)(1)}{(i)(i-1)(...)(2)(1)}=\frac{1}{36}(12-i)(11-i)(10-i)(i+1)(i+2)(i+3)=\frac{1}{36}(-i^6+27i^5-175i^4-495i^3+4136i^2+12348i+7920)$$
Now : $$\sum_{i=0}^n i^6=\frac{n(n+1)(2n+1)(3n^4+6n^3-3n+1)}{42}$$ $$\sum_{i=0}^n i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$$ $$\sum_{i=0}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$ $$\sum_{i=0}^n i^3=(\Sigma_{i=0}^n i)^2$$ $$\sum_{i=0}^n i^2=\frac{n(n+1)(2n+1)}{6}$$ $$\sum_{i=0}^n i=\frac{n(n+1)}{2}$$
So : $$\sum_{i=0}^{n} {12-i\choose3}{3+i\choose3} = \frac{1}{252}(n+1)(-n^6+29n^5-183n^4-1217n^3+8728n^2+48084n+55440)$$
Here, with 9 : $$\sum_{i=0}^{9} {12-i\choose3}{3+i\choose3} = \frac{1}{252}(10)(-531441+1712421-1200663-887193+706968+432756+55440)=\frac{10}{252}288288=11440$$
Or, it can be factored a bit more : $$\frac{1}{252}(n+1)(-n^6+29n^5-183n^4-1217n^3+8728n^2+48084n+55440)=\frac{1}{252}(15-n)(n+2)(n+3)(n+4)(n+1)(n^2-23n+154)$$ which give you again, with 9 : $$\frac{1}{252}(6)(10)(11)(12)(13)(81-207+154)=\frac{1}{21}(6)(10)(11)(13)(28)=\frac{2}{7}(10)(11)(13)(28)=11440$$
servabat
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Hmm, I have thought of just expanding it out. Is there a way without expanding it? I mean without going to $n^6$. – 1-___- Oct 06 '15 at 18:41
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@user2770287 : I doubt so : I don't see any other reduction you could do here on the binomial coefficients there.. – servabat Oct 06 '15 at 18:47
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$$ \sum_{n\geq 3}\binom{n}{3}x^n = \frac{x^3}{(1-x)^4}\tag{1}$$ hence your sum, read as a convolution, is just the coefficient of $x^{15}$ in the square of $(1)$, i.e.:
$$ [x^{15}]\frac{x^6}{(1-x)^8}=[x^{15}]\sum_{n\geq 7}\binom{n}{7}x^{n-1}=\color{red}{\binom{16}{7}}=11440.\tag{2} $$
Jack D'Aurizio
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$$\begin{align}\sum_{i=0}^9\binom {12-i}3\binom {3+i}3 &=\sum_{i=0}^9\binom {12-i}{9-i}\binom {3+i}{i}\\ &=\sum_{i=0}^9(-1)^{9-i}\binom {-3-1}{9-i}(-1)^{i}\binom {-3-1}{i} \qquad&\text{(upper negation)}&\\ &=-\sum_{i=0}^9\binom {-4}{9-i}\binom {-4}i\\ &=-\binom{-8}9\qquad&\text{(Vandermonde)}\\ &=-(-1)^9\binom{17-1}9\qquad&\text{(upper negation)}\\ &=\binom{16}9=\binom {16}7=11440\qquad\blacksquare \end{align}$$
Hypergeometricx
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