Binomial Question (1 Viewer)

[MONKEYjulz]

So i'm having extreme difficulty trying to do this, hope you can help me out =]

(nC0)^2 + (nC1)^2 + (nC2)^2 + ... + (nCn)^2 = (2n)Cn

thanks!

 

[Aerath]

So i'm having extreme difficulty trying to do this, hope you can help me out =]

(nC0)^2 + (nC1)^2 + (nC2)^2 + ... + (nCn)^2 = (2n)Cn

thanks!

Expand (1+x)^n (1+x)^n = (1+x)^2n

Then let x = 1, and compare coefficients.

 

[MONKEYjulz]

Expand (1+x)^n (1+x)^n = (1+x)^2n

Then let x = 1, and compare coefficients.

I tried that. I can't get the compare coefficients bit.

 

[Aerath]

I tried that. I can't get the compare coefficients bit.

Compare the coefficients of the x^n power.

 

[MONKEYjulz]

Compare the coefficients of the x^n power.

I get:
2nCn = [(nCn)(nC0)]+ [(nCn-1)(nC1)]+ [(nCn-2)(nC2)] + ... + [(nC1)(nCn-2)]+ [(nC0)(nCn)]
Is that right?
What's next?

 

[MONKEYjulz]

I get:
2nCn = [(nCn)(nC0)]+ [(nCn-1)(nC1)]+ [(nCn-2)(nC2)] + ... + [(nC1)(nCn-1)]+ [(nC0)(nCn)]
Is that right?
What's next?

typo fixed it

 

[MONKEYjulz]

NVM I GOT IT!
Thanks Aereth, your the best!