That’s a very interesting name you have there.....
Uupps, aren't you afraid to get a lot of problems, which are until today not yet solved?
For example the 3*A + 1 problem...
For example the 3*A + 1 problem...
@ForumMathSolver
Finally! I have a lot of questions, where should I start?
lichess.org/forum/off-topic-discussion/geometry-of-triangle-sizes-within-triangles#1
This maybe? Give the formula for calculating it.
Or prove that (1+2+3...+x)^2 = 1^3+2^3+3^3...+x^3
[I have a solution already, but it is not beautiful...]
Finally! I have a lot of questions, where should I start?
lichess.org/forum/off-topic-discussion/geometry-of-triangle-sizes-within-triangles#1
This maybe? Give the formula for calculating it.
Or prove that (1+2+3...+x)^2 = 1^3+2^3+3^3...+x^3
[I have a solution already, but it is not beautiful...]
Hi teachmewell, do you use mathematical induction for your prove.
@teachmewell said in #4:
> Or prove that (1+2+3...+x)^2 = 1^3+2^3+3^3...+x^3
1 + 2 + 3 + ... + x
= (x + 1) + {(x - 1) + 2} + ((x - 2) + 3) + ... + {(x - x/2) + (x/2 + 1)}
= (x + 1)*(x + 1)*(x + 1)... x/2 times
= x(x + 1)/2 ... (i)
To prove that 1^3 + 2^3 + 3^3 + ... + x^3 = {x(x + 1)/2}^2 [using induction]
Base case: x = 1, LHS = 1^3 = 1 = 1^2 = (1*2/2)^2 = RHS
Inductive hypothesis: assume this is true for x = k, thus:
1^3 + 2^3 + ... k^3 = {k(k+1)/2}^2
Inductive step: Then we have to prove that it is true for x = k + 1:
LHS = 1^3 + 2^3 + ... + k^3 + (k + 1)^3
= {k(k + 1)/2}^2 + (k + 1)^3 = (k + 1)^2[(k^2)/4 + (k + 1_]/2 = (k + 1)^2[k^2 + 4k + 4]/2 = (k + 1)^2(k + 2)^2/2
= [{(k + 1)}{(k+1) + 1}/2]^2 = RHS.
Thus, 1^3 + 2^3 + 3^3 + ... + x^3 = {x(x + 1)/2}^2 ... (ii)
LHS = (1 + 2 + 3 + ... + x)^2
= {x(x + 1)/2}^2 [using (i)]
RHS = 1^3 + 2^3 + ... + x^3
= Σ(i)^3 where i = 1 to x
= {x(x + 1)/2}^2 [using (ii)]
LHS = RHS, Hence proved.
> Or prove that (1+2+3...+x)^2 = 1^3+2^3+3^3...+x^3
1 + 2 + 3 + ... + x
= (x + 1) + {(x - 1) + 2} + ((x - 2) + 3) + ... + {(x - x/2) + (x/2 + 1)}
= (x + 1)*(x + 1)*(x + 1)... x/2 times
= x(x + 1)/2 ... (i)
To prove that 1^3 + 2^3 + 3^3 + ... + x^3 = {x(x + 1)/2}^2 [using induction]
Base case: x = 1, LHS = 1^3 = 1 = 1^2 = (1*2/2)^2 = RHS
Inductive hypothesis: assume this is true for x = k, thus:
1^3 + 2^3 + ... k^3 = {k(k+1)/2}^2
Inductive step: Then we have to prove that it is true for x = k + 1:
LHS = 1^3 + 2^3 + ... + k^3 + (k + 1)^3
= {k(k + 1)/2}^2 + (k + 1)^3 = (k + 1)^2[(k^2)/4 + (k + 1_]/2 = (k + 1)^2[k^2 + 4k + 4]/2 = (k + 1)^2(k + 2)^2/2
= [{(k + 1)}{(k+1) + 1}/2]^2 = RHS.
Thus, 1^3 + 2^3 + 3^3 + ... + x^3 = {x(x + 1)/2}^2 ... (ii)
LHS = (1 + 2 + 3 + ... + x)^2
= {x(x + 1)/2}^2 [using (i)]
RHS = 1^3 + 2^3 + ... + x^3
= Σ(i)^3 where i = 1 to x
= {x(x + 1)/2}^2 [using (ii)]
LHS = RHS, Hence proved.
@e4e5f4exf4nicht said in #5:
> www.spiegel.de/wissenschaft/mensch/collatz-vermutung-deutscher-mathematiker-meldet-loesung-fuer-zahlenraetsel-a-766643.html
I'm sorry bro, I forgot to take my duolingo lessons today :(
> www.spiegel.de/wissenschaft/mensch/collatz-vermutung-deutscher-mathematiker-meldet-loesung-fuer-zahlenraetsel-a-766643.html
I'm sorry bro, I forgot to take my duolingo lessons today :(
Wow, I knew it, but the speed with which you formulated the solution is incredible. you are sure you didn't use AI? No offence but compliment...
but thats cheating....