Prove that 6*Sigma(1 to k) n^2 = (2*k^3) + (3*k^2) + k

J

jk74

New member
Joined
Dec 23, 2018
Messages
2
Hi. Part of the proof for the sum of the first n cubes contains the following identity:

(1) 6*Sigma(1 to k) n^2 = (2*k^3) + (3*k^2) + k
e.g. k=3:
6*(1+4+9) = 2*27 + 3*9 + 3 = 84

I'd like to show the algebraic steps through which the LHS of (1) is equivalent to the RHS.
Any help very gratefully received.

Thanks for reading.
Jon
 
Code: 
n: 1   2   3   4   5    6.....
   1   8  27  64  125  216.....
     7  19  37  61   91 .....
      12  18  24   30.....
         6   6   6 .....
 
jk74 said:
Hi. Part of the proof for the sum of the first n cubes contains the following identity:
(1) 6*Sigma(1 to k) n^2 = (2*k^3) + (3*k^2) + k
e.g. k=3: 6*(1+4+9) = 2*27 + 3*9 + 3 = 84
Click to expand...
You have already proven a base case, \(\displaystyle k=3\)
Now suppose the the sum is true \(\displaystyle k=J>3\) i.e \(\displaystyle 6\sum\limits_{k = 1}^J {{k^2}} = 2{J^3} + 3{J^2} + J\)
Now look at:

\(\displaystyle \begin{align*}6\sum\limits_{k = 1}^{J+1} {{k^2}}&=6\sum\limits_{k = 1}^J {{k^2}}+6(J+1)^2 \\&=[2{J^3} + 3{J^2} + J]+6(J+1)^2\\&=[2{J^3} + 3{J^2} + J]+(6J^2+12J+6)\\&=2{J^3} + 3{J^2} + J+6J^2+12J+6\\&=2J^3+6J^2+6J+2+3J^2+7J+4\\&=[2J^3+6J^2+6J+2]+\{3J^2+6J+3\}+(J+1)\\&=
2[J^3+3J^2+3J+1]+3\{J^2+2J+1\}+(J+1)\\&=2(J+1)^3+3(J+1)^2+(J+1) \end{align*}\)
Thus it is true for \(\displaystyle J+1\)
 
jk74 said:
Hi. Part of the proof for the sum of the first n cubes contains the following identity:

(1) 6*Sigma(1 to k) n^2 = (2*k^3) + (3*k^2) + k
e.g. k=3:
6*(1+4+9) = 2*27 + 3*9 + 3 = 84

I'd like to show the algebraic steps through which the LHS of (1) is equivalent to the RHS.
Any help very gratefully received.

Thanks for reading.
Jon
Click to expand...
I am wondering if you are not aware that

\(\displaystyle \displaystyle \sum_{k=1}^n k^2 = \dfrac{n(n + 1)(2n + 1)}{6}\)

\(\displaystyle \therefore \displaystyle 6 * \sum_{k=1}^n k^2 = n(n + 1)(2n + 1) = (n^2 + n)(2n + 1) = 2n^3 + 3n^2 + n.\)

If you are not aware of the formula for the sum of the first n squares, you will scratch your head about that equation.

EDIT: Of course, this does raise the question of how the formula for the sum of the first n squares is found in the first place.
 
Last edited:
Also interesting is the formula is simply the
square of the sum of numbers formula n(n+1)/2.
 
Last edited:
I am wondering if you are not aware that

[FONT=MathJax_Size2]∑[FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]6[/FONT]∑k=1nk2=n(n+1)(2n+1)6[/FONT]

[FONT=MathJax_AMS]∴[FONT=MathJax_Main]6[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Size2]∑[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main].[/FONT]∴6∗∑k=1nk2=n(n+1)(2n+1)=(n2+n)(2n+1)=2n3+3n2+n.[/FONT]
​
Click to expand...

Thanks to you all. I can see why you are "elite" members. JeffM, you hit the nail on the head. I followed a proof for the n squares sigma formula elsewhere and all is clear. Thank you!
​
 
You must log in or register to reply here.
Top