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@Murch
30 Sep
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on: [Daily puzzle] \sum{n=1}^m n^3 = \left( \sum_{n=1}^m n \right)^2?
science
Yes, it's true. I suspect that m is supposed to be a natural number greater than or equal to 1.
Assuming it's true for any one value in the sequence, we can prove that it's true for all succeeding values per induction.
To go from m to m+1, we show that both sides grow the same:
\sum{n=1}^{m+1} n^3 = \left( \sum_{n=1}^{m+1} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = \left( m+1+ \sum_{n=1}^{m} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + 2*(m+1)*\left(\sum_{n=1}^{m} n\right) + \left( \sum_{n=1}^{m} n \right)^2
Use Gauss sum: (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + 2*(m+1)
m
(m+1)/2 + \left( \sum_{n=1}^{m} n \right)^2
Simplify: (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + m*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2
1
x+nx=(1+n)x: (m+1)^3 + \sum{n=1}^m n^3 = 1
(m+1)^2 + m*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = (1+m)*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2 (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^3 + \left( \sum_{n=1}^{m} n \right)^2
Deduct (m+1)^3 from both sides: \sum{n=1}^m n^3 = \left( \sum_{n=1}^m n \right)^2
For m = 1: sum(1^3) = (sum(1))^2
It's true for m=1, and we have shown that if it's true for any m, it's also true for m+1. QED.
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@south_korea_ln
OP
30 Sep
Another one by induction...
We do need
MathJax support
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(m+1)^3 + \sum{n=1}^m n^3 = \left( m+1+ \sum_{n=1}^{m} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + 2*(m+1)*\left(\sum_{n=1}^{m} n\right) + \left( \sum_{n=1}^{m} n \right)^2
1x+nx=(1+n)x: (m+1)^3 + \sum{n=1}^m n^3 = 1(m+1)^2 + m*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = (1+m)*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2 (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^3 + \left( \sum_{n=1}^{m} n \right)^2