The cubes of the natural numbers are grouped as 1^3 , (2^3 , 3^3 ), (4^3 , 5^3 , 6^3 ), ...., then the sum of the numbers in the nth group is


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Question Number 110015 by ZiYangLee last updated on 26/Aug/20

$The cubes of the natural numbers are$ $grouped as 1^{3}, (2^{3}, 3^{3}), (4^{3}, 5^{3}, 6^{3}), . . . .,$ $then the sum of the numbers in the n th$ $group is$
Commented by Dwaipayan Shikari last updated on 26/Aug/20

$f o r$ $1, (2, 3), (4, 5, 6) . . . .$ $Σ y = Σ (n^{2} - \frac{5 n}{2} + \frac{5}{2}) = \frac{n (n + 1) (2 n + 1)}{6} - \frac{5 n (n + 1)}{4} + \frac{5 n}{2}$
	Answered by Dwaipayan Shikari last updated on 26/Aug/20

	$n t h g r o u p f i r s t t e r m i s y^{3}$ $y △ y △^{2} y$ $1$ $1$ $2 1$ $2$ $4 1$ $3$ $7$ $y = 1 + (n - 1) + \frac{1}{2} (n - 1) (n - 2) = (n^{2} - \frac{5 n}{2} + \frac{5}{2})$ $y^{3} = {(n^{2} - \frac{5 n}{2} + \frac{5}{2})}^{3}$ $\underset{n = 1}{\sum^{n}} {(n^{2} - \frac{5 n}{2} + \frac{5}{2})}^{3} = \frac{1}{672} n (96 n^{6} - 504 n^{5} + 1344 n^{4} - 2205 n^{3} + 2618 n^{2} - 2205 n + 1528)$
	Commented by aurpeyz last updated on 26/Aug/20

	$P l s e x p l a i n h o w y o u g o t t h e n u m b e r s y o u w r o t e$ $i n t h e f i r s t 9 l i n e s a n d h o u y o u a r r i v e d a t$ $n^{2} - \frac{5 n}{2} + \frac{5}{2}$
	Commented by Dwaipayan Shikari last updated on 26/Aug/20

	${N e w t o n}^{'} s f o r w a r d i n t e r p o l a t i o n$ $ϕ (y) = y_{0} + △ y_{0} (n - 1) + △^{2} y_{0} \frac{(n - 1) (n - 2)}{2!} + . . .$ $△ y i s t h e d i f f e r e n c e b e t w e e n s e q u e n c e s$
	Commented by Dwaipayan Shikari last updated on 26/Aug/20
	https://en.m.wikipedia.org/wiki/Interpolation
	Commented by aurpeyz last updated on 27/Aug/20

	$I understand . how come you use 1 2 4 7 ?$
	Answered by mr W last updated on 27/Aug/20

	$1 + 2 + 3 + . . . + (n - 1) = \frac{(n - 1) n}{2}$ $t h e s u m o f t h e n t h g r o u p i s :$ $S_{n} = \underset{k = \frac{(n - 1) n}{2} + 1}{\sum^{\frac{(n - 1) n}{2} + n}} k^{3}$ $= \underset{k = 1}{\sum^{\frac{(n - 1) n}{2} + n}} k^{3} - \underset{k = 1}{\sum^{\frac{(n - 1) n}{2}}} k^{3}$ $= {[\frac{1}{2} (\frac{(n - 1) n}{2} + n) (\frac{(n - 1) n}{2} + n + 1)]}^{2}$ $- {[\frac{1}{2} (\frac{(n - 1) n}{2}) (\frac{(n - 1) n}{2} + 1)]}^{2}$ $= \frac{n^{2} [{(n + 1)}^{2} {(n^{2} + n + 2)}^{2} - {(n - 1)}^{2} {(n^{2} - n + 2)}^{2}}{64}]$ $= \frac{n^{3} (n^{2} + 1) (n^{2} + 3)}{8}$
	Commented by mr W last updated on 26/Aug/20

	$c h e c k :$ $S_{1} = \frac{1^{3} (1^{2} + 1) (1^{2} + 3)}{8} = 1 = 1^{3} \Rightarrow o k$ $S_{2} = \frac{2^{3} (2^{2} + 1) (2^{2} + 3)}{8} = 35 = 2^{3} + 3^{3} \Rightarrow o k$ $S_{3} = \frac{3^{3} (3^{2} + 1) (3^{2} + 3)}{8} = 405 = 4^{3} + 5^{3} + 6^{3} \Rightarrow o k$ $S_{4} = \frac{4^{3} (4^{2} + 1) (4^{2} + 3)}{8} = 2584 = 7^{3} + 8^{3} + 9^{3} + 10^{3} \Rightarrow o k$
	Commented by ZiYangLee last updated on 27/Aug/20

	$Love your solution the most ==$
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