-2
$\begingroup$

Question: How can we prove that $$\displaystyle\left(\sum_{m=1}^{n}{m}\right)^2=\sum_{m=1}^{n}{m^3}$$ holds true?

Edit:- Actually I am new at Stack Exchange and Mathjx. I knew it's geometrical (graphical) representation. But I wanted to do it by induction. After all, (learning from comments) I have posted my own complete solution here.

$\endgroup$
  • $\begingroup$ Do you mean $\left(\sum_{k=1}^nk\right)^2=\sum_{k=1}^nk^3$? $\endgroup$ – Lord Shark the Unknown Jul 4 '18 at 3:46
  • $\begingroup$ @LordSharktheUnknown Yes Please tell me how to solve it. Thanks. $\endgroup$ – Sonu Lamba Jul 4 '18 at 3:47
  • $\begingroup$ If you can compute the LHS, then you can prove both sides are equal by induction on $n$. $\endgroup$ – Lord Shark the Unknown Jul 4 '18 at 3:49
  • $\begingroup$ @LordSharktheUnknown That will be long method. Well I will try to do, but if there is any short method then please tell me. Thanks. $\endgroup$ – Sonu Lamba Jul 4 '18 at 3:53
  • $\begingroup$ More elaboration on your attempt would be appreciated. See also for more tips to inprove your question. $\endgroup$ – Simply Beautiful Art Jul 4 '18 at 3:59
1
$\begingroup$

Using the sum of AP; $$\left(\sum_{m=1}^{n}{m}\right)^2=(1+2+3+...+n)^{2}=\frac{n^{2}(n+1)^{2}}{4}$$

Now by Principle of Mathematical Induction,

for $n=1$,

$$\frac{1^{2}(2)^2}{4}=1$$ so its true for $n=1$,

now for $n=2$, $$\frac{2^{2}(3^{2})}{4}=9=1+2^{3}$$ so its true for $n=2$,

Now assume it holds for $n=k$, and now we can easily show that it is true for $n=k+1$ as follows;

Add $(k+1)^3$, which is $(k+1)^{\mathrm th}$ term of the LHS to both sides; then we get: $$\begin{align*} 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\ &= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\ &= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\ &= \frac{(k+1)^2(k+2)^2}{4}.\\ 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{(k+1)^2(k+2)^2}{4} \end{align*}$$ Hence the proof.

$\endgroup$
  • $\begingroup$ I'd say that the inductive step, which you skipped with we can easily show, is the most relevant part of the proof. And it is a bit unfortunate to say $n=k$, since you have already used the variable $k$ in a different meaning. $\endgroup$ – Martin Sleziak Jul 4 '18 at 13:59
  • $\begingroup$ In any case, if you checked the linked question, you have probably seen that there are other posts on the site which give an inductive proof: Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction. $\endgroup$ – Martin Sleziak Jul 4 '18 at 14:00
  • $\begingroup$ I think now (after edit) it is correct. $\endgroup$ – Sonu Lamba Jul 4 '18 at 15:52
  • $\begingroup$ Yet, the proof seems to be correct. (Almost the same as in this answer.) $\endgroup$ – Martin Sleziak Jul 4 '18 at 16:07
  • $\begingroup$ Actually I am new at stack exchange and Mathjx. I will improve slowly but surly. Thanks. $\endgroup$ – Sonu Lamba Jul 4 '18 at 16:11
0
$\begingroup$

Using the common sum, $$(1+2+3+...+n)^{2}=\frac{n^{2}(n+1)^{2}}{4}$$

Let us try induction, starting with $n=1$,

$$\frac{1^{2}(2)^2}{4}=1$$ so its true for $n=1$,

let us check for $n=2$, $$\frac{2^{2}(3^{2})}{4}=9=1+2^{3}$$ so its true for $n=2$,

Now assume it holds for $n$, can you show it holds for $n+1$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.