Sum the first n^3 how provide with using mathematical induction?

Question

I having next formula and need solving with helps mathematical induction.

$\sum_{i=1}^n i^3 = \biggl(\frac{n(n+1)}{2}\biggl)^2$

My next steps in solving the task:

a)I check the validity of this assertion for n = 1

$\sum_{i=1}^1 1^3 = \biggl(\frac{1(1+1)}{2}\biggl)^2$ = $1$

b)Next, I try to prove the following assertion for n + 1

$\sum_{i=1}^n i^3 = \biggl(\frac{n(n+1)}{2}\biggl)^2$ = $\sum_{i=1}^n i^3$ + $(i + 1)^3$ = $\biggl(\frac{n(n+1)}{2}\biggl)^2$ + $(n+1)^3$ = $\biggl(\frac{n(n+1)}{2}\biggl)^2$ + $(n+1)^3$ = $\biggl(\frac{n(n+1)}{2}\biggl)^2$ + $(n+1)^3$

c)I right solving ? Or have problems ?

Answer 1

You need to show that $$ \sum_{i=1}^{n+1} i^3 = \left( \frac{(n+1)(n+2)}{2}\right)^2, $$

using as an hypothesis that

$$ \sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2}\right)^2. $$

So, \begin{align*} \sum_{i=1}^{n+1} i^3=& \sum_{i=1}^n i^3 + (n+1)^3= \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3\\ =& (n+1)^2\left( \frac{n^2}{4}+n+1\right)=\frac{(n+1)^2}{4}(n^2+4n+4)\\ =&\frac{(n+1)^2(n+2)^2}{4} = \left( \frac{(n+1)(n+2)}{2}\right)^2 \end{align*}

Answer 2

Your induction step makes no sense; random indices seem to be coming in when not referenced, and you have not made an induction hypothesis, which is key to the proof. It would help you to write more words and focus less on expressing everything symbolically.

An induction proof has three main steps:

• Base Case (which you seem to have no issue with)
• Induction Hypothesis (assume the proposition holds for some $n=k$ where $k$ is larger than the base case)
• Induction Step (verify, on the premise that your hypothesis holds, that it implies that $k+1$ holds too)

Thus, here, you assume that the $n=k$ case holds, i.e.

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

Now, we want to verify that

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

This is easy to verify since, pulling out the $(k+1)^{th}$ term from the sum,

$$\sum_{i=1}^{k+1} i^3 = (k+1)^3 + \sum_{i=1}^{k} i^3$$

Applying our induction assumption to the summation on the right, which we assume to hold, we have now that

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

From here, it's a matter of making algebraic manipulations to show that

$$(k+1)^3 + \left( \frac{k(k+1)}{2} \right)^2 = \left( \frac{(k+1)(k+2)}{2} \right)^2$$

as desired. I'll leave the process of doing so to you.