# Thread: Proof on the summation of a finite series

1. ## Proof on the summation of a finite series

Need help on the proof of $\sum^{n}_{r=1}r^2=\frac{1}{6}n(n+1)(2n+1)$

The series of the squares of the first n natural numbers , ie

$\sum^{n}_{r=1}r^2=1^2+2^2+3^2+...+(n-1)^2+n^2$

I know this identity is needed to complete the proof :

$
(r+1)^3=r^3+3r^3+3r+1

$

From here onwards , i am not sure what to do ..

Not sure how to proof this as well :

$
\sum^{n}_{r=1}r^3=\frac{1}{4}n^2(n+1)^2
$

2. Hi

There are several ways to prove that $\sum^{n}_{r=1}r^2=\frac{1}{6}n(n+1)(2n+1)$

One is induction but obviously the beginning of your proof suggests another one.

Write the equality from (r+1) down to 1
$(r+1)^3=r^3+3r^2+3r+1$
$r^3=(r-1)^3+3(r-1)^2+3(r-1)+1$
etc ...
$2^3=1^3+3\cdot 1^2+3\cdot 1+1$
$1^3=0^3+3\cdot 0^2+3\cdot 0+1$

Then sum all these equalities and simplify, some terms being on both sides

3. ## Re :

My book says :

By using this identity

$(r+1)^3=r^3+3r^2+3r+1$

By adding the terms of the identity one by one with values of r from r=n to r=1 , we get

$\sum^{n}_{r=1}[(r+1)^3-r^3]=3\sum^{n}_{r=1}r^2+3\sum^{n}_{r=1}r+\sum^{n}_{r=1 }1$

The LHS side of this identity can be written as :

$[(n+1)^3-n^3]+[n^3-(n-1)^3]+...+(3^3-2^3)+(2^3-1^3)=(n+1)^3-1$

MY question is why is
$[(n+1)^3-n^3]+[n^3-(n-1)^3]+...+(3^3-2^3)+(2^3-1^3)=(n+1)^3-1$

$[(n+1)^3-n^3]+[n^3-(n-1)^3]+...+(3^3-2^3)+(2^3-1^3)=(n+1)^3-1$