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-

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}), \dots .,

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) \dots .

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!} + \dots

△ 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 + \dots + (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 ==