I have to prove the following using induction:
$1^3 + 2^3 + 3^3 + ... + n^3 = \dfrac{n^2(n + 1)^2}{4} $
I understand the base case: we just replace n with 1 and we end up with 1 = 1, so $n=1$ is true.
At the induction step, we have to show that $n = k + 1$ is true.
I looked at the solution and it says:
$1^3 + 2^3 + 3^3 + ... + k^3 + (k+1)^3= \dfrac{(k+1)^2(k + 2)^2}{4} $
I don't understand the left part of this equation. Shouldn't it be:
$1^3 + 2^3 + 3^3 + ... + (k+1)^3$
instead? since we just replace $n$ with $k+1$ ? For me, it is not clear at the moment why we add $k^3$.
You are not replacing n with k+1, you need to assume true for n=k, so write it out with k as the variable instead of n. Then once you have done that you need to show true for k+1. k+1 will include everything in k + k+1. just like if you were at n=2 the lefthand side would read 1 + 2^2 then if it were n=3 the lefthand side would read 1 + 2^2 + 3^3. try adding (k+1)^3 to both sides and then by expanding everything and multiplying (k+1)^3 by 4/4 you should get (k^4 + 6k^3 +13 k^2 + 12k + 4)/4 or something like that. try to factorise again so you have (k^2 + 4k + 4)/4 multiplied by (k^2 + 2k +1)/4 which simplifies to ((k+1)^2(k+2)^2))/4