find the value of 1^2+1 +2^2+2+3^2+3 ---- n^2+n
Answers
Step-by-step explanation:
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!
and in case of math, it is derivations that help you remember well!