Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$

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?

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.

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

protected by Jyrki LahtonenMay 19 '18 at 17:31

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