Do we just cube everything?

Algebra Level 2

Fill in the blank:

$\begin{array} { l l l l l l l l l l l } 1 & + & 2 & + & 3 & + & \cdots & + & n & = &300 \\ 1^3 & + & 2^3 & + & 3^3 & + & \cdots & + & n^3 & = &\text{\_\_\_\_\_\_\_}. \end{array}$

300^4

300^2

300^5

300^3

3 solutions

Md Zuhair
Apr 18, 2017

We know $S_1 =1+2+3 + \cdots n = \dfrac{n(n+1)}{2}$

and $S_2=1^3+2^3+3^3+ \cdots n^3 = \dfrac{n^2(n+1)^2}{4}$

$\implies (S_1)^2=(S_2)$

Now $S_1= 300$ in question so

$\boxed{S_2=300^2}$

Bonus: It can be proven that $1+2+3+\cdots + n = \frac12 n(n+1)$ via the arithmetic progression sum formula. But how do we prove that $1^3+2^3+3^3+\cdots + n^3 = \frac14 n^2(n+1)^2$ ?

Pi Han Goh - 4 years, 1 month ago

I'd go with induction. We know that it's true for n = 1, so then we assume it's true for some n and see if that implies it's true for (n+1). I'm too spent to actually write out all the steps right now.

Denton Young - 4 years, 1 month ago

Yeah, we can prove it by induction. But ever wondered how people derived this formula in the first place? How do we know that sum of cubes can be expressed as that polynomial? Sure we can confirm it, but how was it derived?

Pi Han Goh - 4 years, 1 month ago

@Pi Han Goh – I'm guessing that someone was idly fooling around one day and decided to see what would happen if they added up 1 term in that series, 2 terms in that series, 3 terms in that series... and then discovered that if they added N terms, the sum was the square of the Nth triangular number.

Denton Young - 4 years, 1 month ago

@Denton Young – Nope. Think about $\displaystyle \sum_{j=1}^n [(j+1)^4 - j^4]$ and telescoping sum .

Pi Han Goh - 4 years ago

Munem Shahriar
Jul 28, 2018

We know that, $1+ 2+ 3 + 4 + \ldots + n = \dfrac{n(n+1)}2$ . Now,

$\begin{aligned} \dfrac{n(n+1)}2 & = 300 \\ \Rightarrow n^2 + n & = 600 \\ \Rightarrow n^2 +n - 600 & = 0 \\ \Rightarrow n^2 + 25n-24n -600 & =0\\ \Rightarrow n(n+25) - 24(n+25) & = 0 \\ \Rightarrow (n+25)(n-24) & = 0. \\ \end{aligned}$

We get $n = 24$ . Now,

$\begin{aligned} 1^3 + 2^3 + 3^3 + \ldots + n^3 & = \dfrac{n^2(n+1)^2}4 \\ & = \dfrac{24^2(24 + 1)^2}4 \\ & = \dfrac{4^3 \times 3^2 \times 25^2}4 \\ & = 4^2 \times 3^2 \times 5^2 \times 5^2 \\ &= (4 \times 3 \times 5 \times 5 )^2 \\& = \boxed{300^2}. \\ \end{aligned}$

Great. You can actually simplify your last few steps to $\cdots = 4^2 \times 3^2\times 5^2\times5^2 = (4\times3\times5\times5)^2 = 300^2$ .

Pi Han Goh - 2 years, 10 months ago

I have updated my solution. Thanks.

Munem Shahriar - 2 years, 10 months ago

Zach Abueg
Apr 18, 2017

Use the identity $\displaystyle \sum_{i \ = \ 1}^{n} n^3 = \bigg( \sum_{i \ = \ 1}^{n} n \bigg)^2$

$\displaystyle 1 + 2 + 3\ + ... + \ n = \sum_{i \ = \ 1}^{n} n = 300$

$\displaystyle \sum_{i \ = \ 1}^{n} n^3 = (300)^2$

Bonus: Prove that the only pair of integer solutions $(m,n)$ satisfying $1^m + 2^m + \cdots + j^m = (1^n + 2^n + \cdots + j^n)^p$ for some integer $p>0$ is $(m,n) = (3,1)$ .

Pi Han Goh - 4 years, 1 month ago

How do we tackle this sir? I have the knowledge of the Faulhaber's formula but I think it's not enough to solve such a problem.

Tapas Mazumdar - 4 years, 1 month ago

Here you go

Pi Han Goh - 4 years, 1 month ago

Do we just cube everything?

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