Are they related?

Reply to Chung Kevin on the remarks/questions to Evan Harahap.

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First we will proof:

1. $\sum_{k=1}^n k = \frac{1}{2}n \cdot (n+1)$

2. $\sum_{k=1}^n k^3 = (\frac{1}{2}n \cdot (n+1))^2 = (\sum_{k=1}^n k)^2$

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Proof of (1)

Let $S = \sum_{k=1}^n k = 1 + 2 + 3 + \ldots + (n-1) + n$

and writing the terms backwards we get:

$S = \sum_{k=1}^n k = n + (n-1) + (n-2) + \ldots + 2 + 1$

Summing both identities of $S$ term by term results in:

$2S = (1 + n) + (2 + (n-1)) + (3 + (n-2)) + \ldots + (n + 1)$

$= (n+1) + (n+1) + \ldots + (n+1)$

Because we have $n$ terms (each one being $(n+1)$ ) on the righthand side, we have:

$2S = n \cdot (n+1) \Leftrightarrow S = \frac{1}{2} n \cdot (n+1)$

N.B. We could also proof (1) by induction.

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Proof of (2)

Let $P(n): \sum_{k=1}^n k^3 = (\sum_{k=1}^n k)^2$

$n = 1 \Rightarrow \sum_{k=1}^1 k^3 = 1^3 = 1$ and $(\sum_{k=1}^1 k)^2 = (1)^2 = 1$

Assume $P(n)$ is true. Then:

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

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

$(\frac{1}{2}n \cdot (n+1))^2 + (n+1)^3 =$

$(\frac{1}{2})^2 \cdot n^2 \cdot (n+1)^2 + (n+1)^3 =$

$(\frac{1}{2})^2 \cdot n^2 \cdot (n+1)^2 + (\frac{1}{2})^2 \cdot 4 \cdot (n+1)^3 =$

$(\frac{1}{2})^2 \cdot (n+1)^2 \cdot (n^2 + 4(n+1)) =$

$(\frac{1}{2})^2 \cdot (n+1)^2 \cdot (n+2)^2 =$

$(\frac{1}{2} (n+1) (n+2))^2 = P(n+1)$

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Now for the stated problem:

$1 + 2 + \ldots + n = 300 \Leftrightarrow$

$\frac{1}{2} n (n+1) = 300 \Leftrightarrow$

$n (n+1) - 600 = 0 \Leftrightarrow$

$n^2 + n - 600 = 0 \Leftrightarrow$

$(n+25)(n-24) = 0 \Leftrightarrow$

$n=24$ (because $n > 0$ )

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So a solution exists and $n = 24$ gives:

$1^3 + 2^3 + 3^3 + \ldots + 24^3 =$

$\sum_{k=1}^{24} k^3 =$

$(\sum_{k=1}^{24} k)^2 =$

$(\frac{1}{2}\cdot 24 \cdot (24+1))^2 = 300^2 = 90000$

( 1 + 2 + 3 +... + n) = k for any positive integer n and k, so ( 1^3 + 2^3 + 3^3 +... +n^3) = k^2.

Lets not complicate it... Lets just take the first 3 terms... 1+2+3=6 ; 1^3+2^3+3^3=36 now to verify the same squared relationship consider first 2 terms... 1+2=3 and 1^3+2^3=9 Yay ! that means answer is getting squared...so square of 30 gives us 90000 ! Peace out !

1+2+3+......n = n(n+1)/2 ,
1^3 + ........ n^3 = { n(n+1)/2 }^2 ,
given n(n+1)/2 = 300 ,
therefore 1^3+ .........+n^3 = 300^2 ,
which is equal to 90000 ,
in this way they are related

I really don't have a solution, I just input all the terms in the calculator and found 90000. first, I tried 1+2+3+. . . +n, trial and error until I got the 300, where n = 24. then I cube all those terms, from 1 to 24 using my calculator (and the computer's calculator to arrive at the that value)until I got the 90000 and answered the problem. After answering it, I revealed those three solutions and amazed in their solutions. I tried it in my calculator and it really worked. thank you for those, I've learned.

what had I learned? It is not stopping to solve any problems. be it a mathematics or real life one. Maybe they had their easy way, that theirs are so quick and simple as that, but the Important thing is I tried my own and learned from it. (:

*good thing the sum of the terms is only small-- 300. if it's that large like for example, 3,000,000 or that of the avogadro's number, 6.022x10^23 I think, all my life I can't find the answer. hehehe. But I tried.!

The way i looked at it is like this1^3+ 2^3 +3^3.....+n^3 is the same as (1+2+3....+n)^3. Because of this, we only need to cube the solution to this huge sequence, 300. 300^3=90000