1

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 ?

Boujozo
  • 125
  • 1
    What is the induction hypothesis? Is it not $$\sum\limits_{i=1}^{n} i^3 = \biggl(\frac {n(n+1)} {2}\biggl)^2?$$ – little o Apr 05 '19 at 08:07
  • Maybe try an easier induction problem first, to get used to induction and manipulating series, since it seems that's part of where the problem is, so better to practice first with easier algebra. For example, you could try the following problems, using induction: $$\begin{align} 1) \quad \sum_{i=1}^{n} i &= \frac{n(n+1)}{2} \ 2) \quad \sum_{i=1}^{n} i^2 &= \frac{n(n+1)(2n+1)}{6}.\end{align}$$ – Minus One-Twelfth Apr 05 '19 at 08:11
  • Thank you very much. I need to dig a little algebraic manipulations. – Boujozo Apr 05 '19 at 08:22
  • And again, thank you very much for explaining how to do with algebraic manipulations. – Boujozo Apr 05 '19 at 11:58
  • I know the solution to this hypothesis, but I didn’t know why it was that way, so by examining a more complicated example, I could understand more or less what mathematical induction is. But I still decide her that you offered me and put my decision here to you. – Boujozo Apr 05 '19 at 12:03

2 Answers2

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.

PrincessEev
  • 43,815
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*}

PierreCarre
  • 20,974
  • 1
  • 18
  • 34