prove by pmi 1^2 + 2^2 + 3^2 +....+n^2 > n^3/3
Answers
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!