a,b,c=1,2,3,...,n find Σ


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Question Number 87313 by mr W last updated on 03/Apr/20

$a, b, c = 1, 2, 3, . . ., n$ $f i n d \sum_{a \neq b \neq c} a b c$
Commented by mr W last updated on 04/Apr/20

$y e s, y o u a r e r i g h t .$
Commented by mr W last updated on 04/Apr/20

$f o r e x a m p l e n = 4$ $\sum_{a \neq b \neq c} a b c = 1 \times 2 \times 3 + 1 \times 2 \times 4 + 1 \times 3 \times 4 + 2 \times 3 \times 4 = 50$
Commented by mr W last updated on 03/Apr/20

$t o c h o o s e t h r e e d i f f e r e n t n u m b e r s$ $f r o m n n u m b e r s t h e r e a r e C_{3}^{n}$ $c o m b i n a t i o n s . f r o m e a c h c o m b i n a t i o n$ $w e g e t t h e p r o d u c t o f t h e t h r e e n u m b e r s .$ $n o w i t i s t o f i n d t h e s u m o f a l l t h e s e$ $p r o d u c t s .$
Commented by MJS last updated on 04/Apr/20

$sorry but for n = 4 we have$ $1 \times 2 \times 3 + 1 \times 2 \times 4 + 1 \times 3 \times 4 + 2 \times 3 \times 4 = 50$
	Answered by mr W last updated on 04/Apr/20

	$S = Σ a b c = (\underset{a = 1}{\sum^{n}} a) (\underset{b = 1}{\sum^{n}} b) (\underset{c = 1}{\sum^{n}} c) = {[\frac{n (n + 1)}{2}]}^{3}$ $S_{1} = \sum_{a = b = c} a b c = \underset{a = 1}{\sum^{n}} a^{3} = \frac{n^{2} {(n + 1)}^{2}}{4}$ $S_{2} = \sum_{a, b = c \neq a} a b c = 3 \underset{b = 1}{\sum^{n}} [(\underset{a = 1}{\sum^{n}} a - b) (b^{2})]$ $= 3 \underset{b = 1}{\sum^{n}} [(\underset{a = 1}{\sum^{n}} a) b^{2} - b^{3}]$ $= 3 [(\underset{a = 1}{\sum^{n}} a) (\underset{b = 1}{\sum^{n}} b^{2}) - (\underset{b = 1}{\sum^{n}} b^{3})]$ $= 3 [\frac{n (n + 1)}{2} \times \frac{n (n + 1) (2 n + 1)}{6} - \frac{n^{2} {(n + 1)}^{2}}{4}]$ $= \frac{(n - 1) n^{2} {(n + 1)}^{2}}{2}$ $S_{3} = \sum_{a \neq b \neq c} a b c = \frac{1}{3!} (S - S_{1} - S_{2})$ $= \frac{1}{6} [\frac{n^{3} {(n + 1)}^{3}}{8} - \frac{n^{2} {(n + 1)}^{2}}{4} - \frac{(n - 1) n^{2} {(n + 1)}^{2}}{2}]$ $= \frac{(n - 2) (n - 1) n^{2} {(n + 1)}^{2}}{48}$ $e x a m p l e : n = 4$ $S_{3} = \frac{2 \times 3 \times 4^{2} \times 5^{2}}{48} = 50$ $1 \times 2 \times 3 + 1 \times 2 \times 4 + 1 \times 3 \times 4 + 2 \times 3 \times 3$ $= 50$ $e x a m p l e : n = 5$ $S_{3} = \frac{3 \times 4 \times 5^{2} \times 6^{2}}{48} = 225$ $1 \times 2 \times 3 + 1 \times 2 \times 4 + 1 \times 2 \times 5 + 1 \times 3 \times 4 + 1 \times 3 \times 5 + 1 \times 4 \times 5$ $+ 2 \times 3 \times 4 + 2 \times 3 \times 5$ $+ 3 \times 4 \times 5$ $= 225$
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