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I'm an eight-grader and I need help to answer this math problem (homework).

Problem:

Calculate $$\frac{1^2+2^2+3^2+4^2+...+1000^2}{1^3+2^3+3^3+4^3+...+1000^3}$$

Attempt:

I know how to calculate the quadratic sum using formula from here: Combination of quadratic and arithmetic series but how to calculate the cubic sum? How to calculate the series without using calculator? Is there any intuitive way like previous answer? Please help me. Grazie!

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We have identities for sums of powers like these. In particular:

$$1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$$

$$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$

The rest is just a bit of arithmetic.

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  • $\begingroup$ Similar to theage, you might find it helpful to look at en.wikipedia.org/wiki/Faulhaber%27s_formula, which contains a few more common cases in the examples section. $\endgroup$ – Jason Knapp Jul 24 '14 at 12:58
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    $\begingroup$ The general cases may be a bit scary for an eighth-grader, however. $\endgroup$ – theage Jul 24 '14 at 12:59
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    $\begingroup$ Thanks for the answer but how do I know the formula for cubic sum is $\frac{n^2(n+1)^2}{4}$? $\endgroup$ – L Lawliet Jul 24 '14 at 13:04
  • $\begingroup$ When coming up with power sums like these, it's often very difficult to come up with an elementary intuition. If you've seen proofs by induction, that's the standard way of verifying the identity. $\endgroup$ – theage Jul 24 '14 at 13:07
  • $\begingroup$ Not really a hint :) $\endgroup$ – Thomas Andrews Jul 24 '14 at 13:16
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It is a quadraic sequence as 1,4,9,16,25,36 ........//

Its first term (a)=1 1st difference (d)=4-1=3 2nd difference i.e. constant difference(c)=2

There is a sum formula for any quadraic equation which is

Sn=n/6[(n-1)3d+(n-1)(n-2)C]+an This is for any quadraic even for sum of squares.

I have sum formula for cubic sequence.Let "b" be 1st difference "c" be 2nd difference and "d" be 3rd diffrrence .Also "a" is first term."n" is no. of term.Put all value in below formula except "n" to get sum formula of any cubic sequence.

sum formula

Sn=n/24*[(n-1)12b+(n-1)(n-2)4c+(n-1)(n-2)(n-3)d]+an

Both formula are 100% working.

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  • $\begingroup$ It is a quadraic sequence as 1,4,9,16,25,36 ........// Its first term (a)=1 1st difference (d)=4-1=3 2nd difference i.e. constant difference(c)=2 There is a sum formula for any quadraic equation which is Sn=n/6*[(n-1)3d+(n-1)(n-2)c]+an Putting above value in this formula,we get Sn=n/6*[(n-1)3*3+(n^2-3n+2)2]+1n =n/6[9n-9+2n^2-6n+4]+n =n/6[2n^2+3n-5]+n =n/6[2n^2+3n-5+6] =n/6[2n^2+3n+1] =n/6(2n+1)(n+1) Which is final answer.There is also sum formula for cubic and even quartic sequence. $\endgroup$ – Santosh kurmi May 19 '18 at 17:34

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