MaxWong

42!=1 405 006 117 752 879 898 543 142 606 244 511 569 936 384 000 000 000

Or you mean \(nC2=\frac{1}{4[n+2C2]}\)?

I will solve this one for you

\(nC2=\frac{1}{4(n+2C2)}\)

\(4(n+1)=\frac{((2!)(n-2)!)}{n!}\)

\(4(n+1)=\frac{2}{(n)(n-1)}\)

\(2(n+1)(n-1)(n)=1\)

\(2n^3-2n-1=0\)

We will use the cubic equation.

First \(\Delta_0=0^2-3(2)(-2)=0+12=12\)

Then\(\Delta_1=2(0)^3-9(2)(0)(-2)+27(2)^2(-1)=0+0-108=-108\)

\(C=\sqrt[3]{\frac{\sqrt{(\Delta_1)^2-4(\Delta_0)^3+\Delta_1}}{2}}=\sqrt[3]{\frac{\sqrt{(-108)^2-4(12)^3-108}}{2}}=\frac{\sqrt[6]{4644}}{\sqrt[3]{2}}\)

\(u=\frac{-1+\sqrt3i}{2}\)

\(u^1C=\frac{(\sqrt{3}i-1)(\sqrt[6]{4644})}{\sqrt[3]{16}}\)

\(u^2C=(\frac{\sqrt3i-1}{2})^2(\frac{\sqrt[6]{4644}}{\sqrt[3]{2}})=(\frac{-1-\sqrt3i}{2})(\frac{\sqrt[6]{4644}}{\sqrt[3]{2}})=-\frac{(1+\sqrt3i)(\sqrt[6]{4644})}{\sqrt[3]{16}}\)

\(u^3C=((\frac{-1+\sqrt3i}{2})^3)(\frac{\sqrt[6]{4644}}{\sqrt[3]{2}})=\frac{\sqrt[6]{4644}}{\sqrt[3]{2}}\)

\(n_1=\frac{0+\frac{(\sqrt3i-1)(\sqrt[6]{4644})}{\sqrt[3]{16}}+\frac{12}{\frac{(\sqrt3i-1)(\sqrt[6]{4644})}{\sqrt[3]{16}}}}{6}\)

This is definitely not a natural number XD

\(n_2=\frac{0-\frac{(1+\sqrt3i)(\sqrt[6]{4644})}{\sqrt[3]{16}}+\frac{12}{\frac{(1+\sqrt3i)(\sqrt[6]{4644})}{\sqrt[3]{16}}}}{6}\)

This is still not a natural number XD

\(n_3=\frac{\frac{\sqrt[6]{4644}}{\sqrt[3]{2}}+\frac{12}{\frac{\sqrt[6]{4644}}{\sqrt[3]{2}}}}{6}\)

This is not a natural number :(

LOL This no natural solution.

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

This equals \(\frac{2}{3}\)and it is impossible to obtain its standard form(I call it scientific notation because that's what the teachers said in my school)

\(nC2=\frac{n+2C2}{4}\)

\(4(\frac{n!}{(2)[(n-2)!]})=n+1\)

\(2n(n-1)=n+1\)

\(2n^2-2n=n+1\)

\(2n^2-3n-1=0\)

Using quad. formula,

\(n = {-(-3) \pm \sqrt{(-3)^2-4(2)(-1)} \over 2(2)}\)

\(n=\frac{3+\sqrt{17}}{4}\) or \(n=\frac{3-\sqrt{17}}{4}\)

There are no natural number solutions but some real number solutions.

In this example:

The opposite of 20° is 65, and the hypotenuse is x

Therefore

\(sin 20°= \frac{65}{x}\)

\(x = \frac{65}{sin20°}\)

Which is about 190.05

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

x is the solution of quadratic equation.

Do you mean this?

https://epsstore.ti.com/OA_HTML/csksxvm.jsp?nSetId=76424

1+2+......+n=\(\frac{n(1+n)}{2}\)

1+(1+a)+(1+2a)+......+(1+na)=\(\frac{[1+(1+na)](x)}{2}=\frac{(x)(2+na)}{2}\)

where \(x=n+1\)

Nope.

If the triangle is right......

the length of hypotenuse would be \(\sqrt{7^2+10^2}\) using pythagoras thm.

which would be approx. 12.2

15 > 12.2

Therefore the triangle is obtuse.