Binomials query (1 Viewer)

[namburger]

http://i237.photobucket.com/albums/ff3/Namburger/Binomials.jpg
http://i237.photobucket.com/albums/ff3/Namburger/Binomials1.jpg
http://i237.photobucket.com/albums/ff3/Namburger/Binomials2.jpg

A few i couldnt doooo . Thanks in advance

 

[azn-drew]

1.a)consider n(x+1)^(n-1)

n(x+1)^(n-1)=n[(n-1C0)+(n-1C1)x+...+(n-1Cn-2)x^(n-2)+(n-1Cn-1)x^(n-1)]
let x=1,
n(2^(n-1))=n[(n-1C0)+(n-1C1)+...+(n-1Cn-2)+(n-1Cn-1)]
now, notice that n-1Cn-1=1, and (n-1C0=1)
then,
n(2^(n-1))=n[(n-1C1)+(n-1C2)...+(n-1Cn-2)] + 2n
n(2^(n-1))-2n=n[(n-1C1)+(n-1C2)...+(n-1Cn-2)]
n[(2^(n-1))-2]=n[(n-1C1)+(n-1C2)...+(n-1Cn-2)]

then expand RHS and you get your solution.
For part b), you use trial and error for certain n values and pick the lowest value that is >20000. i think it was 11 or 12? i cant remember.

2. i) x^n(1+x)^n(1+1/x)^n=(1+x)^n(x(1+1/x))^n
x^n(1+x)^n(1+1/x)^n=(1+x)^n(x+1)^n
x^n(1+x)^n(1+1/x)^n=(1+x)^2n
sorry if that's hard to read, the point is that A^n x B^n=(AB)^n,
hence you use that axiom to collect x^n and (1+1/x)^n
ii) this one was a toughie; it took me a while. basically, you notice the RHS and see that they have taken the coefficient of the nth term, so you do the same for the LHS. to do that, you use the result from part (i).
x^n(1+x)^n(1+1/x)^n=[1+x+x^2+...+x^n][(n0)+(n1)x+(n2)x^2+...+x^n][(n0)+(n1)x^-1+(n2)x^-2+...+x^-n] , (nr)=(nCr)
and this also =(1+x)^2n
so,
[1+x+x^2+...+x^n][(n0)+(n1)x+(n2)x^2+...+x^n][(n0)+(n1)x^-1+(n2)x^-2+...+x^-n]=(1+x)^2n
now,, the thing to nitice is that the last 2 set of square brackets can cancel out, leaving the first square bracket to multiply it by x^n. for example, (n1)x*(n1)x^-1 cancel out, so youre left with (n1)^2, then that multiplies with x^1 so you get a coefficient for x^1 of (n1)^2.
for this step, all you really need to state is "taking the coefficient of the nth term"
then you will get,
(n0)(n0)+(n1)(n1)+...+(nCn)(nCn)=(2nCn)
now, (n0)=1,
therefore,
1+(n1)^2+(n2)^2+...+(nCn)^2=(2nCn)

For question 3, or the 3rd image, were there any questions leading up to it? it seems strange that it would just spring up like this. =S

 

[Trebla]

For the last one. Consider the expansion of (x + 1)¹⁰ + (x - 1)¹⁰
= 2¹⁰C₀ + 2¹⁰C₂x² + 2¹⁰C₄x⁴ + 2¹⁰C₆x⁶ + 2¹⁰C₈x⁸ + 2¹⁰C₁₀x¹⁰

Now simply let x = 3:
LHS = 4¹⁰ + 2¹⁰
= 2²⁰ + 2¹⁰
= 2¹⁰(2¹⁰ + 1)
RHS = 2{1 + ¹⁰C₂3² + ¹⁰C₄3⁴ + ¹⁰C₆3⁶ + ¹⁰C₈3⁸ + 3¹⁰}

Now, divide both sides by 2 and the result follows.

 

[azn-drew]

nice. =]

 

[henry08]

I am screwed if any of those questions were asked.