Question

The vallu of 1+2+3+...........+n2

Answer 1

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$<i>$
I have wondered how the closed form for the sum of squares for the first n natural numbers was derived. Given the formula for the sum

1^2+2^2+….+n^2= n(n+1)(2n+1)/6

I learned to prove its correctness using mathematical induction. However, I never understood how the formula was derived in the first place! Knuth writes in The Art of Computer Programming, Volume 1, Third Edition, pp. 32 [Addison-Wesley, June 2009]:

{Most textbooks would simply state those formulas, and prove them by induction. Induction is, of course, a perfectly valid procedure; but it does not give any insight into how on earth a person would ever have dreamed the formula up in the first place, except by some lucky guess.}

While randomly browsing the web, I fortunately stumbled upon a bulletin message that gave a simple derivation of the closed form, as follows:

Let S be the sum of squares of the first n natural numbers, such that

S=1^2+2^2+….+n^2

Our aim is to derive a closed form formula for S in terms of n.

We have (x+1)^3=(x+1)(x^2+2x+1)

implies (x+1)^3=x^3+3x^2+3x+1

Using this, we write the following sequence of equations where

1≤x≤n :

(1+1)^3=1^3+3.1^2+3.1+1

(2+1)^3=2^3+3.2^2+3.2+1

(n+1)^3=n^3+3.n^2+3.n+1

If we add up each of the five columns separately, we get

(1+1)^3+(2+1)^3+…….+(n+1)^3=1^3+2^3+….+n^3+3(1^2+2^2+….+n^2)+3(1+2+….+n)+(1+1+1….+1)

After simplification and substitution, we get

2^3+3^3+…..(n+1)^3= 1^3+2^3+….+n^3+ 3s + 3n(n+1)/2 +n

which, after subtracting the sum of cubes, gives us

(n+1)^3= 1+3s + 3n(n+1)/2 +n

implies 3s= (n+1)^3- 3n(n+1)/2 -n-1

implies 6s= 2n^3+3n^2+n

finally giving us

s=n(n+1)(2n+1)/6.

Hope you find it useful!!

I strongly believe that learning a problem from its root cause will help us more!

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