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Prove that: $(\sum_{i=1}^n i)^2$ = $\sum_{i=1}^n i^3$

I can use the fact that $\sum_{i=1}^n i$ = n(n+1)/2 after the inductive hypothesis is invoked. I'm not sure where to start, I would usually break down one side but there isn't usually two sums, so I'm not sure.

user21820
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JanoyCresva
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  • Is it true for $n=1$? Yes? Now start with the left side as though we went one further than $n$, i.e. to $n+1$ instead. Okay, so what does the left side look like? If we square the summation in the inside? Remember that $\sum\limits_{i=1}^{n+1}i = n+1+\sum\limits_{i=1}^n i$ – JMoravitz Feb 26 '17 at 04:12
  • I have the base case, but I'm confused how you would square the sum on the left side. – JanoyCresva Feb 26 '17 at 04:16
  • Well, you can do it a couple different ways. Either by doing $(\sum\limits_{i=1}^{n+1} i )^2 = (n+1+\sum\limits_{i=1}^n i)^2 = n(n+1+\sum\limits_{i=1}^n i) + 1(n+1+\sum\limits_{i=1}^n i)+(\sum\limits_{i=1}^n i)(n+1+\sum\limits_{i=1}^n i) = \dots$ or easier, use the hint they gave you and say $(\sum\limits_{i=1}^{n+1} i)^2 = ((n+1)(n+2)/2)^2$ – JMoravitz Feb 26 '17 at 04:18

4 Answers4

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Check its true for $n = 1$: LHS = 1 = RHS, hence true

Assume true for $n$:

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

Note that the LHS $= \frac{n^{2}}{4}(n+1)^{2}$

Now we must prove true for $n+1$:

Required to prove $(\sum_{i=1}^{n+1}i)^{2} = \sum_{i=1}^{n+1}i^{3}$

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

hence LHS = RHS, hence true for $n+1$, hence by induction true for $n \in \mathbb{Z}^{+}$

mrnovice
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If $n = 1$, the hypothesis is true.
Assume that the hypothesis is true for $n =k$ i.e.
$(\sum_{i=1}^{k} i)^2 = \sum_{i=1}^{k} i^3$.
Now try to prove that the hypothesis is true if $n =k+1$.

Isha Tarte
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$$\Big(\sum_{i=1}^{n+1}i\Big)^2=\Big(\sum_{i=1}^ni+(n+1)\Big)^2=\Big(\sum_{i=1}^ni\Big)^2+2(n+1)\sum_{i=1}^ni+(n+1)^2=$$ $$\sum_{i=1}^ni^3+n(n+1)^2+(n+1)^2=\sum_{i=1}^ni^3+(n+1)^3=\sum_{i=1}^{n+1}i^3$$

Shaun_the_Post
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$$n=1 \to (\sum_{i=1}^1 i)^2=\sum_{i=1}^1 i^3 \to1=1\\ (***)n=k \to (\sum_{i=1}^k i)^2=\sum_{i=1}^k i^3\\ n=k+1 \to (\sum_{i=1}^{k+1}i)^2=\sum_{i=1}^{k+1} i^3\\ (\sum_{i=1}^{k+1}i)^2=(\sum_{i=1}^{k}i+(k+1))^2=(\sum_{i=1}^{k}i)^2+(k+1)^2+2(k+1)(\sum_{i=1}^{k}i)\to (***)\\= \sum_{i=1}^k i^3+(k+1)^2+2(k+1)(\sum_{i=1}^{k}i)=\\ \sum_{i=1}^k i^3+(k+1)^2+2(k+1)(\dfrac{k(k+1)}{2})=\\ \sum_{i=1}^k i^3+(k+1)^2+(k+1)(k+1)k=\\ \sum_{i=1}^k i^3+(k+1)^2(1+k)=\\ \sum_{i=1}^k i^3+(k+1)^3=\\ \sum_{i=1}^{k+1} i^3$$

Khosrotash
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