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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 nth group is

$$\mathrm{The}\:\mathrm{cubes}\:\mathrm{of}\:\mathrm{the}\:\mathrm{natural}\:\mathrm{numbers}\:\mathrm{are} \\ $$$$\mathrm{grouped}\:\mathrm{as}\:\mathrm{1}^{\mathrm{3}} ,\:\left(\mathrm{2}^{\mathrm{3}} ,\:\mathrm{3}^{\mathrm{3}} \right),\:\left(\mathrm{4}^{\mathrm{3}} ,\:\mathrm{5}^{\mathrm{3}} ,\:\mathrm{6}^{\mathrm{3}} \right),\:...., \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{the}\:{n}\mathrm{th} \\ $$$$\mathrm{group}\:\mathrm{is} \\ $$

Commented by Dwaipayan Shikari last updated on 26/Aug/20

for 1,(2,3),(4,5,6).... Σy=Σ(n^2 −((5n)/2)+(5/2))=((n(n+1)(2n+1))/6)−((5n(n+1))/4)+((5n)/2)

$${for} \\ $$$$\mathrm{1},\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{4},\mathrm{5},\mathrm{6}\right).... \\ $$$$\Sigma{y}=\Sigma\left({n}^{\mathrm{2}} −\frac{\mathrm{5}{n}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{2}}\right)=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\frac{\mathrm{5}{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}}+\frac{\mathrm{5}{n}}{\mathrm{2}} \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 26/Aug/20

nth group first term is y^3 y △y△^2 y 1 1 2 1 2 4 1 3 7 y=1+(n−1)+(1/2)(n−1)(n−2)=(n^2 −((5n)/2)+(5/2)) y^3 =(n^2 −((5n)/2)+(5/2))^3 Σ_(n=1) ^n (n^2 −((5n)/2)+(5/2))^3 =(1/(672))n(96n^6 −504n^5 +1344n^4 −2205n^3 +2618n^2 −2205n+1528)

$${nth}\:{group}\:{first}\:{term}\:{is}\:{y}^{\mathrm{3}} \\ $$$${y}\:\bigtriangleup{y}\bigtriangleup^{\mathrm{2}} {y} \\ $$$$\mathrm{1} \\ $$$$\:\:\:\:\:\:\mathrm{1} \\ $$$$\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\mathrm{2} \\ $$$$\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{3} \\ $$$$\mathrm{7} \\ $$$${y}=\mathrm{1}+\left({n}−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)=\left({n}^{\mathrm{2}} −\frac{\mathrm{5}{n}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$${y}^{\mathrm{3}} =\left({n}^{\mathrm{2}} −\frac{\mathrm{5}{n}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({n}^{\mathrm{2}} −\frac{\mathrm{5}{n}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{672}}{n}\left(\mathrm{96}{n}^{\mathrm{6}} −\mathrm{504}{n}^{\mathrm{5}} +\mathrm{1344}{n}^{\mathrm{4}} −\mathrm{2205}{n}^{\mathrm{3}} +\mathrm{2618}{n}^{\mathrm{2}} −\mathrm{2205}{n}+\mathrm{1528}\right) \\ $$

Commented by aurpeyz last updated on 26/Aug/20

Pls explain how you got the numbers you wrote in the first 9 lines and hou you arrived at n^2 −((5n)/2)+(5/2)

$${Pls}\:{explain}\:{how}\:{you}\:{got}\:{the}\:{numbers}\:{you}\:{wrote}\: \\ $$$${in}\:{the}\:{first}\:\mathrm{9}\:{lines}\:{and}\:{hou}\:{you}\:{arrived}\:{at}\: \\ $$$${n}^{\mathrm{2}} −\frac{\mathrm{5}{n}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Commented by Dwaipayan Shikari last updated on 26/Aug/20

Newton′s forward interpolation φ(y)=y_0 +△y_0 (n−1)+△^2 y_0 (((n−1)(n−2))/(2!))+... △y is the difference between sequences

$${Newton}'{s}\:{forward}\:{interpolation} \\ $$$$\phi\left({y}\right)={y}_{\mathrm{0}} +\bigtriangleup{y}_{\mathrm{0}} \left({n}−\mathrm{1}\right)+\bigtriangleup^{\mathrm{2}} {y}_{\mathrm{0}} \frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{2}!}+... \\ $$$$\bigtriangleup{y}\:{is}\:{the}\:{difference}\:{between}\:{sequences} \\ $$

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?

$$\mathrm{I}\:\mathrm{understand}.\:\mathrm{how}\:\mathrm{come}\:\mathrm{you}\:\mathrm{use}\:\mathrm{1}\:\mathrm{2}\:\mathrm{4}\:\mathrm{7}? \\ $$

Answered by mr W last updated on 27/Aug/20

1+2+3+...+(n−1)=(((n−1)n)/2) the sum of the nth group is: S_n =Σ_(k=(((n−1)n)/2)+1) ^((((n−1)n)/2)+n) k^3 =Σ_(k=1) ^((((n−1)n)/2)+n) k^3 −Σ_(k=1) ^(((n−1)n)/2) k^3 =[(1/2)((((n−1)n)/2)+n)((((n−1)n)/2)+n+1)]^2 −[(1/2)((((n−1)n)/2))((((n−1)n)/2)+1)]^2 =((n^2 [(n+1)^2 (n^2 +n+2)^2 −(n−1)^2 (n^2 −n+2)^2 )/(64))] =((n^3 (n^2 +1)(n^2 +3))/8)

$$\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\left({n}−\mathrm{1}\right)=\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}} \\ $$$${the}\:{sum}\:{of}\:{the}\:{nth}\:{group}\:{is}: \\ $$$${S}_{{n}} =\underset{{k}=\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}+\mathrm{1}} {\overset{\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}+{n}} {\sum}}{k}^{\mathrm{3}} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}+{n}} {\sum}}{k}^{\mathrm{3}} −\underset{{k}=\mathrm{1}} {\overset{\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}} {\sum}}{k}^{\mathrm{3}} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}+{n}\right)\left(\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}+{n}+\mathrm{1}\right)\right]^{\mathrm{2}} \\ $$$$−\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}\right)\left(\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}+\mathrm{1}\right)\right]^{\mathrm{2}} \\ $$$$\left.=\frac{{n}^{\mathrm{2}} \left[\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}^{\mathrm{2}} +{n}+\mathrm{2}\right)^{\mathrm{2}} −\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}^{\mathrm{2}} −{n}+\mathrm{2}\right)^{\mathrm{2}} \right.}{\mathrm{64}}\right] \\ $$$$=\frac{{n}^{\mathrm{3}} \left({n}^{\mathrm{2}} +\mathrm{1}\right)\left({n}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{8}} \\ $$

Commented by mr W last updated on 26/Aug/20

check: S_1 =((1^3 (1^2 +1)(1^2 +3))/8)=1=1^3 ⇒ok S_2 =((2^3 (2^2 +1)(2^2 +3))/8)=35=2^3 +3^3 ⇒ok S_3 =((3^3 (3^2 +1)(3^2 +3))/8)=405=4^3 +5^3 +6^3 ⇒ok S_4 =((4^3 (4^2 +1)(4^2 +3))/8)=2584=7^3 +8^3 +9^3 +10^3 ⇒ok

$${check}: \\ $$$${S}_{\mathrm{1}} =\frac{\mathrm{1}^{\mathrm{3}} \left(\mathrm{1}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{8}}=\mathrm{1}=\mathrm{1}^{\mathrm{3}} \:\Rightarrow{ok} \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{2}^{\mathrm{3}} \left(\mathrm{2}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{8}}=\mathrm{35}=\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} \:\Rightarrow{ok} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{3}^{\mathrm{3}} \left(\mathrm{3}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{8}}=\mathrm{405}=\mathrm{4}^{\mathrm{3}} +\mathrm{5}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} \:\Rightarrow{ok} \\ $$$${S}_{\mathrm{4}} =\frac{\mathrm{4}^{\mathrm{3}} \left(\mathrm{4}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{4}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{8}}=\mathrm{2584}=\mathrm{7}^{\mathrm{3}} +\mathrm{8}^{\mathrm{3}} +\mathrm{9}^{\mathrm{3}} +\mathrm{10}^{\mathrm{3}} \:\Rightarrow{ok} \\ $$

Commented by ZiYangLee last updated on 27/Aug/20

Love your solution the most ==

$$\mathrm{Love}\:\mathrm{your}\:\mathrm{solution}\:\mathrm{the}\:\mathrm{most}\:== \\ $$

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