If ^{n }C_{r}, denotes the coefficient of x^{r} in the expansion of (1 + x)^{ n}, prove that:

^{n }C_{r} + 2(^{n }C_{r+1}) + ^{n }C_{r+2} = ^{n+2 }C_{r+2}

OldTimer
Jan 10, 2018

#1**0 **

Hi Old timer,

This is not a proof, ... I just wanted to convince myself that your statement was true.

I considered Pascals triangle which gives the coefficients nCr

Just considering how the pattern works it can be deduced that

^{n}C_{r}+^{n}C_{r+1 }= ^{n+1}C_{r+1 } and ^{n}C_{r+1}+^{n}C_{r+2} = ^{n+1}C_{r+2}

_{and}

^{n+1}C_{r+1 } + ^{n+1}C_{r+2 = }^{n+2}C_{r+2}

_{so}

^{n}C_{r}+^{n}C_{r+1 }+ ^{n}C_{r+1}+^{n}C_{r+2} = ^{n+2}C_{r+2}

_{hence}

^{n}C_{r}+ 2^{n}C_{r+1 }+^{n}C_{r+2} = ^{n+2}C_{r+2}

To prove it I would probably use the fact that nCr = n! /(r! * (n-r)! )

But I will admit that I have not worked through this proof.

Melody
Jan 10, 2018