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MaxWong

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MaxWong  13 ene 2019
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Or you mean nC2=14[n+2C2]?

I will solve this one for you

nC2=14(n+2C2)

4(n+1)=((2!)(n2)!)n!

4(n+1)=2(n)(n1)

2(n+1)(n1)(n)=1

2n32n1=0

We will use the cubic equation.

First Δ0=023(2)(2)=0+12=12

ThenΔ1=2(0)39(2)(0)(2)+27(2)2(1)=0+0108=108

C=3(Δ1)24(Δ0)3+Δ12=3(108)24(12)31082=6464432

u=1+3i2

u1C=(3i1)(64644)316

u2C=(3i12)2(6464432)=(13i2)(6464432)=(1+3i)(64644)316

u3C=((1+3i2)3)(6464432)=6464432

n1=0+(3i1)(64644)316+12(3i1)(64644)3166

This is definitely not a natural number XD

n2=0(1+3i)(64644)316+12(1+3i)(64644)3166

This is still not a natural number XD

n3=6464432+1264644326

This is not a natural number :(

LOL This no natural solution.

Sorry i have gone a bit crazy solving that cubic eq.

12 jun 2016