200 sats \ 1 reply \ @CruncherDefi 30 Sep \ on: [Daily puzzle] \sum{n=1}^m n^3 = \left( \sum_{n=1}^m n \right)^2? science
Yes, assuming it's true, it's true for both m and m-1. Lets substract both-sides of equations for m and m-1 and we get:
m^3 = ( 1+2+3+...m)^2 - (1+2+3...(m-1))^2
Rth side is two arithmetic series. We can simplify:
m^3 = ( (1+m) * (m) / 2 )^2 - ( (1+m-1) * (m-1) / 2) ) ^ 2
m^3 = ( (m+1) * m / 2 )^2 - ( (m-1) * m / 2 )^2
Divide both sides by (m / 2)^2
4m = ( (m+1) )^2 - ( (m-1) )^2
4m = ( (m+1 + m-1) ) * ( (m+1-m+1) )
4m = ( 2m ) * ( 2 )
4m = 4m
A proof by induction, I like it...
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