- #1
dleacock
hey everyone,
I recently bought the the Dover Series book on Number Theory, and the 2nd example on page 5 asks your to prove
[tex]1^3 + 2^3 + 3^3 ... + n^3 = (1 + 2 + 3...)^2 [/tex]
Now, we've already proved that [tex]S_n = \frac{n(n+1)}{2}[/tex]
So here's how I proved it...
[tex]
(S_n)^2 = (\frac{n(n+1)}{2})^2 [/tex]
before we proved how [tex] S_k+1 = S_k + (k + 1) [/tex]
Which lead me to...
[tex] \frac{1}(k+1)^2((k+1)+1)^2{4} + (k+1)^2 [/tex]
[tex] = \sqrt{\frac{(k+1)^2((k+1)+1)^2}{4} + (k+1)^2} [/tex]
[tex] = \frac{(k+1)((k+1)+1)}{2} + (k+1) [/tex]
therefor...
[tex] S_k+1 = Sk + (k+1) [/tex]
Now I'm worried that I didnt really solve anything. I'm totally knew at this, and I'm open to criticism and help, just be kind :)
(ps.. this is my first time posting formulas, hopefully I did it right)
Thanks
dleacock
I recently bought the the Dover Series book on Number Theory, and the 2nd example on page 5 asks your to prove
[tex]1^3 + 2^3 + 3^3 ... + n^3 = (1 + 2 + 3...)^2 [/tex]
Now, we've already proved that [tex]S_n = \frac{n(n+1)}{2}[/tex]
So here's how I proved it...
[tex]
(S_n)^2 = (\frac{n(n+1)}{2})^2 [/tex]
before we proved how [tex] S_k+1 = S_k + (k + 1) [/tex]
Which lead me to...
[tex] \frac{1}(k+1)^2((k+1)+1)^2{4} + (k+1)^2 [/tex]
[tex] = \sqrt{\frac{(k+1)^2((k+1)+1)^2}{4} + (k+1)^2} [/tex]
[tex] = \frac{(k+1)((k+1)+1)}{2} + (k+1) [/tex]
therefor...
[tex] S_k+1 = Sk + (k+1) [/tex]
Now I'm worried that I didnt really solve anything. I'm totally knew at this, and I'm open to criticism and help, just be kind :)
(ps.. this is my first time posting formulas, hopefully I did it right)
Thanks
dleacock