Using the principle of mathematical induction, prove that $$\sum_{k=1}^{n} k^3 = (\frac{n(n+1)}{2})^2;\ \ \ \ n \geq 1 $$

Question

Using the principle of mathematical induction, prove that $$\sum_{k=1}^{n} k^3  = (\frac{n(n+1)}{2})^2;\ \ \ \ n \geq 1 $$

sasogeek · Answer

typical question to expect from agd

perl · Answer

base case is k=1 
then show if it is true for n=k , it is true for k+1

anonymous · Answer

but I'm stuck at the part where you simplify it after the k+1 part to show it is true :(

anonymous · Answer

I have to reduce the following to something trivial: $$(\frac{(m+1)(m+2)}{2})^2 = (\frac{m(m+1)}{2})^2 + (m+1)^3$$

mathmate · Answer

RHS:
(m(m+1)/2)^2+(m+1)^2
=(m+1)^2[(m/2)^2+(m+1)]
=(m+1)^2[m^2+4m+4]/4
=(m+1)^2[(m+2)^2]/4
=[(m+1)(m+2)/2]^2

anonymous · Answer

wow

anonymous · Answer

hmm.... in the first step, you have (m+1)^2... is that a typo?

mathmate · Answer

Yes, it was.  [just woke up!]