Problem doing induction question

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The discussion revolves around proving the formula 2³+4³+6³+...+(2n)³=2n²(n+1)² using mathematical induction. The initial base case for n=1 is verified correctly. The user assumes the statement holds for n=k and attempts to prove it for n=k+1, but encounters confusion in their algebraic manipulation. They realize they made an error by misrepresenting a term in their calculations. The thread highlights the importance of careful algebraic handling in induction proofs.
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Homework Statement


Prove the following result using mathematical induction:
2³+4³+6³+...+(2n)³=2n²(n+1)² for all n>or=1


Homework Equations





The Attempt at a Solution


n=1:
(2(1))³=2(1)²(2)³
8=8

Assume n=k
2³+4³+6³+...+(2k)³=2k²(k+1)²

n=k+1
2³+4³+6³+...+(2k)³+(2(k+1))³=2(k+1)²(k+2)²
Using assumption
2k²(k+1)²+(2(k+1))³=2(k+1)²(k+2)²
Divide by 2(k+1)²
k²+k+1=(k+2)²
k²+k+1=k²+4k+4

I can't understand where I've gone wrong.
Any help would be greatly appreciated.
 
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Figured it out the second I put it up. Removed a 2³ as a 2.
 
sbsbsbsb said:

Homework Statement


Prove the following result using mathematical induction:
2³+4³+6³+...+(2n)³=2n²(n+1)² for all n>or=1

Homework Equations



The Attempt at a Solution


n=1:
(2(1))³=2(1)²(2)³
8=8

Assume n=k
2³+4³+6³+...+(2k)³=2k²(k+1)²

n=k+1
2³+4³+6³+...+(2k)³+(2(k+1))³=2(k+1)²(k+2)²
Using assumption
The following line looks as if you are assuming the very thing you are to prove.
2k²(k+1)²+(2(k+1))³=2(k+1)²(k+2)²
Divide by 2(k+1)²
k²+k+1=(k+2)²
k²+k+1=k²+4k+4

I can't understand where I've gone wrong.
Any help would be greatly appreciated.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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