Question

Prove the sum of cubes of 1, ... , n equals (n*(n+1)/2)^2

Answer 1

lol truthfully i just memorized the stuff for my exams nver thought of proving it in my school..
i will still try

Answer 2

Here's the LaTex for the equation:

\[1^{3} + 2^{3} + ... + n^{3} = (n(n+1)/2)^{2}\]

Answer 3

you know you use mathematical induction
Check the truth of the statement for 1:
1^3=((1*(1+1))/2)^2=1^2=1. Thus the statement is true for 1
Now, assume that for n, the statement is true
And now, we prove the statement for n+1
Thus, we need to prove
\[1^3+2^3+...+n^3+(n+1)^3=((n+1)(n+2)/2))^2\]
Now, since
\[1^3+2^3+...+n^3=(n(n+1)/2)^2\] is true
then
\[1^3+2^3+...+n^3+(n+1)^3=(n(n+1)/2)^2+(n+1)^3\] is true
Now simplify \[(n(n+1)/2)^2+(n+1)^3\]
Eventually you will get \[((n+1)(n+2)/2)^2\]
Yeah, and so you can conclude that the statement is true for n+1. And so the statement has been proven

Answer 4

lol this is a good method!! but we also have another method which i don't seem to remember!

Answer 5

Thats mathematical induction.. pretty neat!

Answer 6

Thanks Anhhuyalex!

Akshay - If you find the proof without using induction, I'd certainly be interested to see it.