1

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 $k^3$.

Michael
  • 233
  • 2
    Think of an example: 1^3+2^3+3^3+...+6^3. What's the second to last term? i.e the term before 6^3, it's 5^3; so if you want to be explicit we can write 1^3+2^3+3^3+...+5^3+6^3 (both sums are the same, you are not adding anything). – Daniel López Aguayo Apr 25 '20 at 17:20
  • 3
    $1^3 + 2^3 + 3^3 + \dotsb + k^3 + (k + 1)^3$ and $1^3 + 2^3 + 3^3 + \dotsb + (k + 1)^3$ are two ways to write the same thing! Both are "the sum of all the cubes from $1^3$ up to $(k + 1)^3$". (This can be written $\sum_{r = 1}^{k + 1} r^3$). The dots just mean to "complete the obvious pattern", which is technically ambiguous, but the meaning should be clear here. Remember that the inductive step is to prove that "IF the statement is true for $n = k$, THEN is is also true for $n = k + 1$". (cf https://math.stackexchange.com/questions/19485/dominoes-and-induction-or-how-does-induction-work) – Izaak van Dongen Apr 25 '20 at 17:25

1 Answers1

1

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

jacob bradley
  • 186