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I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:

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

I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?

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31 Answers 31

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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

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.

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  • 1
    $\begingroup$ Why is the sum formula true then? $\endgroup$ – Alex Vong May 19 '18 at 17:00
  • $\begingroup$ @AlexVong This formula is made by myself using a long theory of sequence.Combining arithmetic sequence formula it is formed $\endgroup$ – Santosh kurmi May 19 '18 at 17:22
  • $\begingroup$ It is real sum formula of quadraic sequence $\endgroup$ – Santosh kurmi May 19 '18 at 17:26
  • $\begingroup$ @AlexVong It is made using Sum formula of AP $\endgroup$ – Santosh kurmi May 19 '18 at 17:27
  • $\begingroup$ Since the question asked why the formula is true, you may want to add the reason to the answer. $\endgroup$ – Alex Vong May 20 '18 at 5:58

protected by Jyrki Lahtonen May 19 '18 at 17:31

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