Question

(et ( S=sum_{r=1}^{n} r^{2} )
( n^{3}-(n-1)^{3}=3 n^{2}-3 n+1 )
Substituting ( n=1,2,3,4,5 )
[
begin{array}{c}
1^{3}-0^{3}=3 cdot 1^{2}-3 cdot 1+1
2^{3}-1^{3}=3 cdot 2^{2}-3 cdot 2+1
n^{3}-(n-1)^{3}=3 cdot n^{2}-3 cdot n+1
end{array}
]
Aading we get
[
n^{3}-0^{3}=3left(1^{2}+2^{2}+3^{2}+4^{2}+cdots+n^{2}right)-3(1+2+3+n)
]
[
begin{array}{l}
quad+(1+1+1+1 cdots+n)
n^{3}=35-3 frac{n(n+1)}{2}+n
3 S=frac{n(n+1)(2 n+1)}{2}
S=frac{n(n+1)(2 n+1)}{6}
end{array}
]

# What is sum of the square of the first in Natural nubers Tequation with derivation

Solution