Binomials (1 Viewer)

[Hotdog1]

I posted this in the 3 Unit forum, but no one seems to go there so can I interest any of you 13370R2?


Q50
by considering the coefficinets of
1+(1+x)+(1+x)^2+...+(1+x)^n,
prove that [nCr]+[(n-1)Cr]+[(n-2)Cr]+...+[rCr] = [(n+1)C(r+1)]
similarly consdering
x^n+[x^(n-1)](1+x)+[x^(n-2)](1+x)^2+...+(1+x)^n,
find the sum [nCr]+[(n-1)C(r-1)]+[(n-2)C(r-2)]+...+[(n-r-1)C1]

Q51
If P_r = [nCr]*x^r*(1-x)^(n-r),
Show that P_1 + 2P_2 +3P_3 + ... + nP_n =nx

Q52
By considering the coefficients of x^2r (2r < n) in the expansion of (1+x)^n*(1-x)^n and (1-x^2)^n,
show that
[nC(r-1)][nC(r+1)] - [nC(r-2)][nC(r+2)] + ... + (-1)^r*[nC1][nC(2r-1)] - (-1)^r*[nC2r] = (1/2)[nCr][(nCr)-1]

Thanks alot!

 

[underthesun]

For the first question,

It's a GP. The sum of the GP is ((1+x)ⁿ⁺¹ - 1)/x
Now, you find the coefficients of x^r in the long series, and you find the same for the sum. Note that you have to find the coefficient of (n+1)C(r+1), because the value will get divided by x. So you'll need a value of x^r+1 on the top, and you'll end up with what the question wants. It's long working, there's no way i can type it here..

 

[underthesun]

The second question misleads me.. it seems..

I'd have used the expansion of 1ⁿ = (x + (1-x))ⁿ

which gives terms like P_r
where P_r = ⁿC_rx^r(1-x)^n-r.

But when when I want to find that identity, i'm lost..

 

[MyLuv]

Q50:Use GP as UndertheSun:
1st one:GP with n+1 term common ratio = (1+x) ,then equal co. of x^r
2nd one : GP with n+1 term commonratio=(1+x)/x,then equal co. of x^r
Q51:
Consider co. of even power of x (x^2k) eg:X^2,4,6,8...
P_1= - n!/(n-2k)!*(2k-1)!
2kP_2k= n!/(n-2k)!*(2k-1)! ---> P_1+ nP_n=0
similarly,2P_2+ (2k-1)P_(2k-1)=0 , 3P_3 +(2k-2)P_(2k-2)=0...
--> Co.efficient of x^2k =0
Consider co. of odd power of x (x^2k+1) eg:x^3,5...
P_1= n!/(2k)!(n-1-2k)!
2P_2=-n!/(2k-1)!(n-1-2k)!
....
(2k)P_2k= -n!/(2k-1)!(n-1-2k)!
(2k+1)P_2k+1= n!/(2k)!(n-1-2k)!
A=Sum of them={n!*[ 1/(2k)!- 1/(2k-1)!+...-1(2k-1)!+1/(2k)!]}/(n-1-2k)!

make common denominator = 2k!
--->A= {n![2kC0 -2kC1+...-2kC(2k-1)+ 2kC2k]}/(2k)!*(n-1-2k)!
Now, 2kC0-2kC1.....+2kC2k=0 (In bionomial,Sum of Even Co= Odd co.)
---> A=0
---> Co. efficient of x^2k+1 is 0
So, P_1 +2P_2+..+nP_n = kx (since co. of X^2,3,4....=0)
coeficient of x is nx
--> P_1 +2P_2+...+ nP_2=nx
Q52:

x^2r *(2r

<---what U mean???

ps: Btw,what is 1337 ??? On Bnet,ppl use it too but I dont get it

 

[Dumbarse]

1337 is LEET
as in elite
computer geek talk

 

[Hotdog1]

Oops the Html stuffed up, I fixed it now. It was meant to be x^2r (2r < n)

Thanks underthesun and MyLuv for answering.

U can do the first question now, but I'm still somewhat unsure about the second one.

MyLuv: can you explain this step please?

Originally posted by MyLuv
---> Co. efficient of x^2k+1 is 0
So, P_1 +2P_2+..+nP_n = kx (since co. of X^2,3,4....=0)
coeficient of x is nx

Also is this a standard set approach to this type of question? it seems quite complicated

Binomials

 

[MyLuv]

---> Co. efficient of x^2k+1 is 0
So, P_1 +2P_2+..+nP_n = kx (since co. of X^2,3,4....=0)
coeficient of x is nx

P_1 +2P_2+...+ nP_n= k1*x +k2 *x^2+k3*x^3...+kn*x^n
but co . efficient of x^2,3,4,5...=0 (as shown in 2 case above)
---> P_1+2P_2+...+nP_n= kx

Ps: U can make it shorter by just using a general x^k and do the same as the case x^2k+1

 

[MyLuv]

Q52: (1+x)^n*(-x+1)^n = (1-x^2)^n
consider co. of x^2r in LHS:
Co.= nC2r-nC(2r-1)*nC1+...+nC(r+1)*nC(r-1)*(-1)^r-1+ nCr*nCr*(-1)^r+ nC(r-1)*nC(r+1)*(-1)^r+1+...+nC2r*(-1)^2r
Now,
nC2r=nC2r*(-1)^2r
-nC(2r-1)*nC1= nC1*nC(2r-1)*(-1)^(2r-1) since nC(2r-1)=nC1(due to 2nd 'n 2nd last term of (x+1)^n) and nC1 = nC(2r-1) for similar reason
...
nC(r+1)*nC(r-1)*(-1)^r-1=nC(r-1)*nC(r+1)*(-1)^r+1 (similar reason)
only nCr*nCr(-1)^r is single

--->Co. of x^2r in LHS= nCr*nCr*(-1)^r + 2*[nC(r-1)*nC(r+1)*(-1)^(r+1)+....+nC2r*(-1)^2r]

Co. of x^2r in RHS=nCr(-1)r

LHS=RHS--->nCr*nCr*(-1)^r + 2*[nC(r-1)*nC(r+1)*(-1)^(r+1)+....+nC2r*(-1)^2r]=nCr(-1)^r

-->nCr*nCr*(-1)^r- nCr*(-1)^r=-2*[nC(r-1)*nC(r+1)*(-1)^(r+1)+....+nC2r*(-1)^2r]
---> (-1)^r *nCr[ nCr-1]= - 2 * (-1)^r *[ -nC(r-1)*nC(r+1)+...+nC2r*(-1)^r]
--->1/2 nCr[nCr-1] = nC(r-1)*nC(r+1)+...(-1)^r*nC2r
Done

 

[Hotdog1]

Thank you very much, MyLuv