I have a question that's giving me a hard time and im not sure how to go about solving it...
Prove that 1^3 + 2^3 + ... + n^3 = [n(n+1)/2]^2 while n>= 1
im a little lost at what to do after testing it out for n = 1 and proving that true.
Mathematcal Induction
For P(1):
Put n=1
RHS =LHS
Hence our assumption is correct for n=1
Assumption P(K):
Let the above statement be correct for n=k
Hence
1^3 + 2^3 + ... + k^3 = [k(k+1)/2]^2.............................1
Prove it For P(k+1):
So LHS will be
1^3 + 2^3 + ... + k^3 + (k+1)^3
Now for terms till n=k put the values obtained in (1)
[k(k+1)/2]^2 + (k+1)^3
On simplifying the above thing you will get
[(k+1)(k+2)/2]^2
This proves that the statement is correct for n = (k+1) when its correct for n= k
Hence through the principle of mathematical induction we have proved that the statement is correct