Question

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

Answer 1

typical question to expect from agd

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

wow

Answer 7

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

Answer 8

Yes, it was. [just woke up!]