Find the value of : Σ_(i=0) ^∞ Σ_(j=0) ^∞ Σ_(k=0) ^∞ (1/(3^i 3^j 3^k )). case 1: i≠j≠k. case 2: i<j<k.


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Question Number 31743 by rahul 19 last updated on 13/Mar/18

$F i n d t h e v a l u e o f : \underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}} \frac{1}{3^{i} 3^{j} 3^{k}} .$ $c a s e 1 : i \neq j \neq k .$ $c a s e 2 : i < j < k .$
	Answered by mrW2 last updated on 13/Mar/18
	$Σ_(k=0) ^∞ (1/3^k )=1+(1/3)+(1/3^2 )+....=(1/(1−(1/3)))=(3/2) Σ_(i=0) ^∞ Σ_(j=0) ^∞ Σ_(k=0) ^∞ (1/(3^i 3^j 3^k )) =Σ_(i=0) ^∞ Σ_(j=0) ^∞ {(1/(3^i 3^j ))(Σ_(k=0) ^∞ (1/3^k ))} =(3/2)Σ_(i=0) ^∞ Σ_(j=0) ^∞ (1/(3^i 3^j )) =(3/2)Σ_(i=0) ^∞ {(1/3^i )(Σ_(j=0) ^∞ (1/3^j ))} =(3/2)×(3/2)Σ_(i=0) ^∞ (1/3^i ) =(3/2)×(3/2)×(3/2) =((27)/8)$
	$\underset{k = 0}{\sum^{\infty}} \frac{1}{3^{k}} = 1 + \frac{1}{3} + \frac{1}{3^{2}} + . . . . = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}$ $\underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}} \frac{1}{3^{i} 3^{j} 3^{k}}$ $= \underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} {\frac{1}{3^{i} 3^{j}} (\underset{k = 0}{\sum^{\infty}} \frac{1}{3^{k}})}$ $= \frac{3}{2} \underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \frac{1}{3^{i} 3^{j}}$ $= \frac{3}{2} \underset{i = 0}{\sum^{\infty}} {\frac{1}{3^{i}} (\underset{j = 0}{\sum^{\infty}} \frac{1}{3^{j}})}$ $= \frac{3}{2} \times \frac{3}{2} \underset{i = 0}{\sum^{\infty}} \frac{1}{3^{i}}$ $= \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}$ $= \frac{27}{8}$
	Commented bymrW2 last updated on 14/Mar/18

	$\underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}} \frac{1}{3^{i} 3^{j} 3^{k}} = \frac{27}{8}$ $t h i s i s c o r r e c t .$ $i f y o u n e e d \underset{i \neq j \neq k}{\underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}}} \frac{1}{3^{i} 3^{j} 3^{k}} o r$ $\underset{i > j > k}{\underset{i = 0}{\sum^{\infty}} \underset{j = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{\infty}}} \frac{1}{3^{i} 3^{j} 3^{k}} t h e n t h e r e s u l t i s d i f f e r e n t .$
	Commented byrahul 19 last updated on 14/Mar/18

	$B u t w h a t i s t h e d i f f e r e n c e w h e n$ $i \neq j \neq k a n d i < j < k a r e p r e s e n t i n q u e s .$ $(o r i g i n a l q .) a n d w h e n t h a t i s n o t m e n t i o n$ $i n t h e q u e s . (t h e w a y u s o l v e d) .$
	Commented bymrW2 last updated on 14/Mar/18

	$l o o k a t a n e x a m p l e :$ $\underset{i = 1}{\sum^{3}} \underset{j = 1}{\sum^{3}} a^{i} b^{j} = a^{1} (b^{1} + b^{2} + b^{3}) + a^{2} (b^{1} + b^{2} + b^{3}) + a^{3} (b^{1} + b^{2} + b^{3}) = (a^{1} + a^{2} + a^{3}) (b^{1} + b^{2} + b^{3})$ $(i a n d j a r e i n d e p e n d e n t)$ $\underset{i \neq j}{\underset{i = 1}{\sum^{3}} \underset{j = 1}{\sum^{3}}} a^{i} b^{j} = a^{1} (b^{2} + b^{3}) + a^{2} (b^{1} + b^{3}) + a^{3} (b^{1} + b^{2})$ $(i a n d j a r e d e p e n d e n t)$ $\underset{i < j}{\underset{i = 1}{\sum^{3}} \underset{j = 1}{\sum^{3}}} a^{i} b^{j} = a^{1} (b^{2} + b^{3}) + a^{2} b^{3}$ $(i a n d j a r e d e p e n d e n t)$
	Commented byrahul 19 last updated on 14/Mar/18

	$o k s i r, n o w u n d e r s t o o d w h e n i t s i n$ $2 v a r i a b l e s (i, j) .$ $p l s a l s o s h o w s i m i l a r e x a m p l e f o r$ $3 v a r i a b l e (i, j, k) ?$
	Commented byabdo imad last updated on 14/Mar/18

	$t h e r e i s a d f f e r e n c e b e t w e e n \sum_{i j} a_{i} b j a n d \sum_{i, j a n d i \neq j} a_{i} a_{j}$ $a n d t h e t y p e o f t h i s t r i p l e s u m i s d i f f i c u t t o t r e a t$ $a n d i f y o u g i v e i n t e r e s t y o u m u s t t a k e a l o o k t o$ $s o m m a b l e s f a m i l y . . . .$
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