Binomial Problem =O (1 Viewer)

[BeDifferent]

If you could do it, it'll be much appreciated.

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

I obtained that part, but the problem lies in the second part

(n) +2( n ) + ( n ) = (n+2)
(r)-----(r-1)----(r-2)----( r )
Yea, didn't really know how to do that short hand notation for nCk

Whoops editted

 

[Timothy.Siu]

nvm, i'm really confused...is this from a textbook, i dont really understand your typing

 

[lyounamu]

your question is ....um....well.....

i cannot read it and i don't know what you are essentially asking.

 

[dp624]

If you could do it, it'll be much appreciated.

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

I obtained that part, but the problem lies in the second part

(n) +2( n ) + ( n ) = (n+2)
(r)-----(r-1)----(r-2)----( r )
Yea, didn't really know how to do that short hand notation for nCk

So what he is saying is that
(x^n)*[(1+1/x)^n]*[(1+x)^2]=(1+x)^2
but this is actually wrong.... lmao!


ok
so RTP:
nCr + 2nCr-1 + nCr-2 = n+2Cr

so we compare crap to get it.
LHS = (1+x)^n (1+x)^2 = (1+x)^n (1+2x+x^2)
RHS = (1+x)^(n+2)

compare the term of x^r in each

LHS = (nC0 + nC1*x + nC2*x^2 + nC3*x^3... + nCn*x^n) (1+2x+x^2)
RHS = ((n+2)C0 + (n+2)C1*x + (n+2)C2*x^2 + (n+2)C3*x^3... + (n+2)C(n+2)*x^(n+2))

So compare the terms of x^r

from RHS-> (n+2)Cr*x^r

from LHS we group the terms such that the final thingy is ^r.
so we take nCr*x^r*1 + nC(r-1)*x^(r-1)*2x + nC(r-2)*x^(r-2)*x^2 =

soo..

nCr + 2nC(r-1)+ nC(r-2) = (n+2)Cr

as required.
QED