Discrete Maths - Induction

Question

I am having difficulty answering the following question:

Can anyone show me how to solve this? I understand that I should be putting in a + 1 somewhere to simulate the next step, but I'm not sure where.

Thanks!

Answer 1

Induction begins by showing a base case. Here our base case is $n=1$ since we want to prove the statement for all integers $n\geq1$.

Case $n=1$:

Note that $1^3=1$ and that$\frac{1^2}{4}\cdot(1+1)^2=\frac{4}{4}=1$. Since these are the same, the base case holds.

Now we make the induction hypothesis: that there exists an integer $k\geq1$ such that the formula holds. That is

$$1^3+2^3+3^3+\cdots+k^3=\sum_{i=1}^{k}i^3=\frac{k^2}{4}(k+1)^2$$

for some integer $k\geq1$.

Now we take the induction step, to consider the case $n=k+1$.

Case $n=k+1$:

$$1^3+2^3+3^3+\cdots+k^3 +(k+1)^3$$

Now by our induction hypothesis we can replace the sum of the first $k$ cubes with $\frac{k^2}{4}(k+1)^2$.

So

$$1^3+2^3+3^3+\cdots+k^3 +(k+1)^3=\frac{k^2}{4}(k+1)^2+(k+1)^3=\frac{1}{4}(k^2+3k+2)^2=\frac{(k+1)^2}{4}(k+2)^2$$

Which we should note is the formula that we are trying to prove for general $n$, with $k+1$ substituted.

Now, this proves the statement for all integers $n\geq1$. This is because we supposed that it held true for one integer, and showed that this implies it will hold true for the next integer. Since we could repeat this argument indefinitely to show that the formula holds for the next integer and we have an integer as our starting point (the base case), we have shown the formula to be true for all integers $n\geq1$.

Answer 2

$$n=1: 1^3=\frac{1}{4} 2^2=1 \checkmark$$

$$\text{ We suppose that: } 1^3+2^3+ \dots + n^3=\frac{n^2}{4}(n+1)^2$$

$$\text{ It suffices to show that: } 1^3+2^3+ \dots+ n^3+(n+1)^3=\frac{(n+1)^2}{4}(n+2)^2$$

$$1^3+2^3+ \dots+ n^3+(n+1)^3=\frac{n^2}{4}(n+1)^2+(n+1)^3=(n+1)^2(\frac{n^2}{4}+n+1) \\ =(n+1)^2 (\frac{n^2+4n+4}{4})=(n+1)^2 \frac{(n+2)^2}{4}$$

So,we conclude that $1^3+2^3+ \dots + n^3=\frac{n^2}{4}(n+1)^2, \forall n \in \mathbb{N}$