Matematické Fórum

stuart clark

$\displaystyle \sum\sum_{0 \leq i < j < k \leq n}\sum 2017$

kerajs · 15. 11. 2017 09:40 — Editoval kerajs (15. 11. 2017 09:49)

$n>1$

$...=2017 \cdot \sum_{i=0}^{j-1}( \sum_{j=i+1}^{k-1} ( \sum_{k=j+1}^{n} 1)) =2017 \cdot \sum_{l=1}^{n-1}T_l =...$
$T_l$ - trojúhelníkové číslo (triangular number)

a)
$n=2p+1$
$...=2017 \cdot (4+16+36+...+(2p)^2)=2017\cdot 4\cdot (1^2+2^2+3^2+...+p^2)=$
$=2017\cdot 4 \frac{p(p+1)(p+2)}{6}= \frac{2017(n-1)(n+1)(n+3)}{12}$

b)
$n=2p$
.....
.....

check_drummer · 15. 11. 2017 19:18

↑ stuart clark:
Hi, is it $2017.{n+1 \choose 3}$ ? because i,j,k is number of 3 members sets in {0,..,n}

kerajs · 15. 11. 2017 22:01

kerajs napsal(a):
a)
$n=2p+1$
$...=2017 \cdot (4+16+36+...+(2p)^2)=2017\cdot 4\cdot (1^2+2^2+3^2+...+p^2)=$
$=2017\cdot 4 \frac{p(p+1)(p+2)}{6}= \frac{2017(n-1)(n+1)(n+3)}{12}$

$2017\cdot 4\cdot (1^2+2^2+3^2+...+p^2) \not = 2017\cdot 4 \frac{p(p+1)(p+2)}{6}$
Sorry

$2017\cdot 4\cdot (1^2+2^2+3^2+...+p^2) = 2017\cdot 4 \frac{p(p+1)(2p+1)}{6}=$
$=2017\cdot 4 \frac{\frac{n-1}{2}(\frac{n-1}{2}+1)(2\frac{n-1}{2}+1)}{6}=$
$=2017\cdot 4 \frac{\frac{n-1}{2}(\frac{n+1}{2})(n)}{6}=$
$=2017\cdot \frac{(n-1)(n+1)(n)}{6}=2017 {n+1 \choose 3}$

b)
$n=2p$
$...=2017 \cdot (4+16+36+...+(2p-2)^2)+2017\cdot T_{2p-1}=$
$=2017\cdot 4 \frac{(p-1)(p-1+1)(2(p-1)+1)}{6}+2017\frac{(2p-1)2p}{2}=$
$=2017\cdot 4 \frac{(\frac{n}{2}-1)(\frac{n}{2})(n-1)}{6}+2017\frac{(n-1)n}{2}=$
$=2017{n+1 \choose 3}$