calculate Σ_((i,j)∈N) ((i^2 +j^2 )/2^(i+j) )


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Question Number 46731 by maxmathsup by imad last updated on 30/Oct/18

$c a l c u l a t e \sum_{(i, j) \in N} \frac{i^{2} + j^{2}}{2^{i + j}}$
Commented by maxmathsup by imad last updated on 01/Nov/18

$S = \sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} \frac{i^{2}}{2^{i} 2^{j}} + \sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} \frac{j^{2}}{2^{i} 2^{j}}$ $= 2 \sum_{i = 0}^{\infty} \frac{i^{2}}{2^{i}} . \sum_{j = 0}^{\infty} \frac{1}{2^{j}} b u t \sum_{j = 0}^{\infty} \frac{1}{2^{j}} = \frac{1}{1 - \frac{1}{2}} = 2 a n d \sum_{i = 0}^{\infty} \frac{i^{2}}{2^{i}} = w (\frac{1}{2}) w i t h$ $w (x) = \sum_{n = 0}^{\infty} n^{2} x^{n} l e t t a k e ∣ x ∣< 1 w e h a v e \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow$ $\sum_{n = 1}^{\infty} {n x}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow \sum_{n = 1}^{\infty} {n x}^{n} = \frac{x}{{(1 - x)}^{2}} \Rightarrow$ $\sum_{n = 1}^{\infty} n^{2} x^{n - 1} = \frac{{(1 - x)}^{2} - x (- 2) (1 - x)}{{(1 - x)}^{4}} = \frac{1 - x + 2 x}{{(1 - x)}^{3}} = \frac{1 + x}{{(1 - x)}^{3}} \Rightarrow$ $\sum_{n = 1}^{\infty} n^{2} x^{n} = \frac{x + x^{2}}{{(1 - x)}^{3}} \Rightarrow w (\frac{1}{2}) = \frac{\frac{1}{2} + \frac{1}{4}}{{(\frac{1}{2})}^{3}} = 8 . \frac{3}{4} = 6 \Rightarrow$ $S = 2 . 6.2 = 24 \Rightarrow ★ S = 24 ★ .$
	Answered by MrW3 last updated on 01/Nov/18

	$S_{n} = \underset{i = 0}{\sum^{n}} \underset{j = 0}{\sum^{n}} \frac{i^{2} + j^{2}}{2^{i + j}}$ $\frac{i^{2} + j^{2}}{2^{i + j}} = \frac{i^{2} + j^{2}}{2^{i} 2^{j}} = \frac{i^{2}}{2^{i}} \times \frac{1}{2^{j}} + \frac{1}{2^{i}} \times \frac{j^{2}}{2^{j}}$ $S_{n} = \underset{i = 0}{\sum^{n}} \underset{j = 0}{\sum^{n}} \frac{i^{2} + j^{2}}{2^{i + j}} = \underset{i = 0}{\sum^{n}} \frac{i^{2}}{2^{i}} \times \underset{j = 0}{\sum^{n}} \frac{1}{2^{j}} + \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}} \times \underset{j = 0}{\sum^{n}} \frac{j^{2}}{2^{j}}$ $\Rightarrow S_{n} = 2 \underset{i = 0}{\sum^{n}} \frac{i^{2}}{2^{i}} \times \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}}$ $P = \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}} = 2 (1 - \frac{1}{2^{n + 1}})$ $Q = \underset{i = 0}{\sum^{n}} \frac{i}{2^{i}} = \frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \frac{4}{2^{4}} + . . . + \frac{n}{2^{n}}$ $2 Q = 1 + \frac{2}{2} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + . . . + \frac{n}{2^{n - 1}}$ $2 Q = 1 + \frac{1 + 1}{2} + \frac{1 + 2}{2^{2}} + \frac{1 + 3}{2^{3}} + . . . + \frac{1 + n - 1}{2^{n - 1}}$ $2 Q = (1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + . . . + \frac{1}{2^{n - 1}}) + (\frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + . . . + \frac{n - 1}{2^{n - 1}})$ $2 Q = 2 (1 - \frac{1}{2^{n}}) + (\frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + . . . + \frac{n - 1}{2^{n - 1}} + \frac{n}{2^{n}}) - \frac{n}{2^{n}}$ $2 Q = 2 (1 - \frac{1}{2^{n}}) + Q - \frac{n}{2^{n}}$ $\Rightarrow Q = 2 (1 - \frac{1}{2^{n}}) - \frac{n}{2^{n}}$ $\Rightarrow Q = \underset{i = 0}{\sum^{n}} \frac{i}{2^{i}} = 2 (1 - \frac{n + 2}{2^{n + 1}})$ $R = \underset{i = 0}{\sum^{n}} \frac{i^{2}}{2^{i}} = \underset{i = 0}{\sum^{n}} \frac{i^{2} - 1 + 1}{2^{i}} = \underset{i = 0}{\sum^{n}} \frac{i^{2} - 1}{2^{i}} + \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}}$ $= \underset{i = 0}{\sum^{n}} \frac{(i - 1) (i - 1 + 2)}{2^{i}} + \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}}$ $= \underset{i = 0}{\sum^{n}} \frac{{(i - 1)}^{2}}{2^{i}} + \underset{i = 0}{\sum^{n}} \frac{2 (i - 1)}{2^{i}} + \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}}$ $= \underset{i = 0}{\sum^{n}} \frac{{(i - 1)}^{2}}{2^{i}} + 2 \underset{i = 0}{\sum^{n}} \frac{i}{2^{i}} - \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}}$ $= \frac{1}{2} \underset{i = 0}{\sum^{n}} \frac{{(i - 1)}^{2}}{2^{i - 1}} + 2 Q - P$ $= \frac{1}{2} [2 + \frac{1}{2} + \frac{2^{2}}{2^{2}} + \frac{3^{2}}{2^{3}} + . . . + \frac{{(n - 1)}^{2}}{2^{n - 1}}] + 2 Q - P$ $= 1 + \frac{1}{2} [\frac{1}{2} + \frac{2^{2}}{2^{2}} + \frac{3^{2}}{2^{3}} + . . . + \frac{{(n - 1)}^{2}}{2^{n - 1}} + \frac{n^{2}}{2^{n}}] - \frac{n^{2}}{2^{n + 1}} + 2 Q - P$ $\Rightarrow R = 1 + \frac{R}{2} - \frac{n^{2}}{2^{n + 1}} + 2 Q - P$ $\Rightarrow \frac{R}{2} = 1 - \frac{n^{2}}{2^{n + 1}} + 2 \times 2 (1 - \frac{n + 2}{2^{n + 1}}) - 2 (1 - \frac{1}{2^{n + 1}})$ $\Rightarrow \frac{R}{2} = 3 - \frac{n^{2} + 4 n + 6}{2^{n + 1}}$ $\Rightarrow R = \underset{i = 0}{\sum^{n}} \frac{i^{2}}{2^{i}} = 2 (3 - \frac{n^{2} + 4 n + 6}{2^{n + 1}})$ $\Rightarrow S_{n} = 2 \underset{i = 0}{\sum^{n}} \frac{i^{2}}{2^{i}} \times \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}} = 2 R P = 2 \times 2 (3 - \frac{n^{2} + 4 n + 6}{2^{n + 1}}) \times 2 (1 - \frac{1}{2^{n + 1}})$ $\Rightarrow S_{n} = 8 (3 - \frac{n^{2} + 4 n + 6}{2^{n + 1}}) (1 - \frac{1}{2^{n + 1}})$ $\lim_{n \to \infty} S_{n} = 24$
	Commented by prof Abdo imad last updated on 31/Oct/18

	$t h n k y o u s i r .$
	Commented by behi83417@gmail.com last updated on 01/Nov/18

	$s i r m r W 3! y o u a r e a l w a y s t h e b e s t .$
	Commented by MrW3 last updated on 01/Nov/18

	$t h a n k y o u t o o s i r s!$ $I a m t h i n k i n g a b o u t h o w t o c a l c u l a t e$ $Σ \frac{i^{3}}{2^{i}}$ $Σ \frac{i^{4}}{2^{i}}$ $e t c .$ $a n y i d e a ?$ $m y t r y i s t o u s e t h e s a m e m e t h o d a s a b o v e, e . g .$ $i^{3} - {(i - 1)}^{3} = 3 i^{2} - 3 i + 1$ $\Rightarrow i^{3} = {(i - 1)}^{3} + 3 i^{2} - 3 i + 1$
	Commented by MrW3 last updated on 01/Nov/18

	$l e t S_{n} = \underset{i = 0}{\sum^{n}} \underset{j = 0}{\sum^{n}} \frac{i^{3} + j^{3}}{2^{i + j}} = 2 \underset{i = 0}{\sum^{n}} \frac{i^{3}}{2^{i}} \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}}$ $w e h a v e a l r e a d y :$ $P = \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}} = 2 (1 - \frac{1}{2^{n + 1}})$ $Q = \underset{i = 0}{\sum^{n}} \frac{i}{2^{i}} = 2 (1 - \frac{n + 2}{2^{n + 1}})$ $R = \underset{i = 0}{\sum^{n}} \frac{i^{2}}{2^{i}} = 2 (3 - \frac{n^{2} + 4 n + 6}{2^{n + 1}})$ $l e t T = \underset{i = 0}{\sum^{n}} \frac{i^{3}}{2^{i}}$ $s i n c e i^{3} = {(i - 1)}^{3} + 3 i^{2} - 3 i + 1$ $T = \underset{i = 0}{\sum^{n}} \frac{{(i - 1)}^{3} + 3 i^{2} - 3 i + 1}{2^{i}}$ $T = \underset{i = 0}{\sum^{n}} \frac{{(i - 1)}^{3}}{2^{i}} + 3 R - 3 Q + P$ $T = \frac{1}{2} \underset{i = 0}{\sum^{n}} \frac{{(i - 1)}^{3}}{2^{i - 1}} + 3 R - 3 Q + P$ $T = \frac{1}{2} (- 2) + \frac{1}{2} \underset{i = 1}{\sum^{n}} \frac{{(i - 1)}^{3}}{2^{i - 1}} + 3 R - 3 Q + P$ $T = - 1 + \frac{1}{2} \underset{i = 0}{\sum^{n - 1}} \frac{i^{3}}{2^{i}} + 3 R - 3 Q + P$ $T = - 1 + \frac{1}{2} \underset{i = 0}{\sum^{n}} \frac{i^{3}}{2^{i}} - \frac{1}{2} \times \frac{n^{3}}{2^{n}} + 3 R - 3 Q + P$ $T = - 1 + \frac{1}{2} T - \frac{1}{2} \times \frac{n^{3}}{2^{n}} + 3 R - 3 Q + P$ $\frac{T}{2} = - 1 - \frac{1}{2} \times \frac{n^{3}}{2^{n}} + 3 R - 3 Q + P$ $\Rightarrow T = - 2 - \frac{n^{3}}{2^{n}} + 6 R - 6 Q + 2 P$ $\Rightarrow T = - 2 - \frac{n^{3}}{2^{n}} + 12 (3 - \frac{n^{2} + 4 n + 6}{2^{n + 1}}) - 12 (1 - \frac{n + 2}{2^{n + 1}}) + 4 (1 - \frac{1}{2^{n + 1}})$ $\Rightarrow T = - 2 - 2 \times \frac{n^{3}}{2^{n + 1}} + 36 - 12 \times \frac{n^{2} + 4 n + 6}{2^{n + 1}} - 12 + 12 \times \frac{n + 2}{2^{n + 1}} + 4 - 4 \times \frac{1}{2^{n + 1}}$ $\Rightarrow T = 26 - \frac{2 n^{3} + 12 (n^{2} + 4 n + 6 - n - 2) + 4}{2^{n + 1}}$ $\Rightarrow T = \underset{i = 0}{\sum^{n}} \frac{i^{3}}{2^{i}} = 2 (13 - \frac{n^{3} + 6 n^{2} + 18 n + 26}{2^{n + 1}})$ $S_{n} = \underset{i = 0}{\sum^{n}} \underset{j = 0}{\sum^{n}} \frac{i^{3} + j^{3}}{2^{i + j}} = 2 \underset{i = 0}{\sum^{n}} \frac{i^{3}}{2^{i}} \underset{i = 0}{\sum^{n}} \frac{1}{2^{i}} = 2 T P$ $= 2 \times 2 (13 - \frac{n^{3} + 6 n^{2} + 18 n + 26}{2^{n + 1}}) 2 (1 - \frac{1}{2^{n + 1}})$ $\Rightarrow S_{n} = 8 (13 - \frac{n^{3} + 6 n^{2} + 18 n + 26}{2^{n + 1}}) (1 - \frac{1}{2^{n + 1}})$ $\lim_{n \to \infty} S_{n} = 8 \times 13 = 104$
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