Or you mean nC2=14[n+2C2]?
I will solve this one for you
nC2=14(n+2C2)
4(n+1)=((2!)(n−2)!)n!
4(n+1)=2(n)(n−1)
2(n+1)(n−1)(n)=1
2n3−2n−1=0
We will use the cubic equation.
First Δ0=02−3(2)(−2)=0+12=12
ThenΔ1=2(0)3−9(2)(0)(−2)+27(2)2(−1)=0+0−108=−108
C=3√√(Δ1)2−4(Δ0)3+Δ12=3√√(−108)2−4(12)3−1082=6√46443√2
u=−1+√3i2
u1C=(√3i−1)(6√4644)3√16
u2C=(√3i−12)2(6√46443√2)=(−1−√3i2)(6√46443√2)=−(1+√3i)(6√4644)3√16
u3C=((−1+√3i2)3)(6√46443√2)=6√46443√2
n1=0+(√3i−1)(6√4644)3√16+12(√3i−1)(6√4644)3√166
This is definitely not a natural number XD
n2=0−(1+√3i)(6√4644)3√16+12(1+√3i)(6√4644)3√166
This is still not a natural number XD
n3=6√46443√2+126√46443√26
This is not a natural number :(
LOL This no natural solution.
Sorry i have gone a bit crazy solving that cubic eq.